Unit outcomes

By the end of this unit you will be able to:

• Identify and describe vertical projectile motion.
• Identify that gravitational acceleration is independent of the mass of an object.
• Identify, analyse and solve problems on vertical projectile motion using equations of motion.

What you should know

Before you start this unit, make sure you can:

• Describe motion and identify and define the components of motion including:
  • displacement
  • velocity
  • constant acceleration
    Refer to level 2 subject outcome 2.1 units 1 and 2 if you need help with this.
• Analyse and calculate problems dealing with motion in one-dimension using equations of motion. Refer to level 3 subject outcome 2.1 unit 1 if you need help with this.

Introduction

Consider the situation shown in figure 1. A big heavy bowling ball and a super light feather are held at a certain height above the ground. If they are dropped, what happens?

You have probably seen enough objects dropped to know that both the bowling ball and the feather will fall to the ground. But what makes them fall? Do they fall at the same speed? Which one hits the ground first?

Most, people when asked these questions, will respond that the objects fall because of gravity. Gravity ‘pulls’ the objects to the ground. They will go on to say that the bowling ball will fall faster and hit the ground first. This is after all our experience of the real world.

But, if the bowling ball falls faster and hits the ground first and if the force of gravity acting on the bowling ball and the feather is the same, then the bowling ball must fall faster because it is heavier. Its extra mass must cause it to be pulled towards the earth faster. Right? Well, no!

Gravity

Many people believe that heavier objects fall faster because their experience shows them that bowling balls fall faster than feathers. But the only reason that bowling balls fall faster than feathers is because of air resistance. If you drop two similarly sized and shaped objects of very different masses from the same height, they will hit the ground at the same time!

If we make the objects the same shape and size, then we minimise the relative effects of air resistance on the two objects. Why not try this yourself now?

Activity 1.1: Investigate falling

Time required: 5 minutes

What you need:

• a small ball (e.g. a bouncy ball or tennis ball)
• a few pieces of paper
• sticky tape

What to do:

Take a few pieces of paper and scrunch them up into a ball of about the same size as the real ball that you have. Make sure that each ball is the same size. Wrap a few pieces of tape around your paper ball to make sure it keeps its shape and size.

Now, hold both balls above your head and drop them. Do this a few times to make sure your results are repeatable.

What did you find?

You should have found, if both balls were the same shape and size, that they both hit the ground at the same time. It does not matter if one is heavier than the other. If we remove the relative effects of air resistance, we can see that both balls fall at the same rate and hit the ground at the same time.

To show that it does not matter what the mass of an object is, it will always fall at the same rate, watch the video called Brian Cox visits the world’s biggest vacuum to see how, when all the air from a room is removed and the effects of air resistance are removed completely, a bowling ball and a feather fall at exactly the same rate and hit the ground at exactly the same time (see figure 2).

Figure 2: A bowling ball and a feather fall at the same rate

We know from Newton’s Law of Universal Gravitation that an object in the earth’s gravitational field experiences a force pulling it towards the centre of the earth. If this is the only force acting on the object (i.e. we ignore or remove air resistance), then the object will accelerate towards the centre of the earth. We can show that the value of this acceleration is [latex]\scriptsize 9.8~\text{m.s}^{-2}[/latex]. All objects, irrespective of their mass, experience the same rate of acceleration and, therefore, fall at the same rate.

For almost all everyday purposes, we can treat the acceleration due to gravity as a constant [latex]\scriptsize 9.8~\text{m.s}^{-2}[/latex]. It is only when one gets far away from the earth, that this ‘gravitational force’ changes.

Motion due to gravity

In the next activity, we will investigate the motion of a ball as it is thrown vertically up into the air.

Activity 1.2: Investigate balls in the air

Time required: 10 minutes

What you need:

• a ball

What to do:

Go outside where the air is as still as possible and throw a ball vertically up into the air. Watch carefully what happens to the ball between the moment it leaves your hand and when it hits the ground.

We will assign upwards as the positive direction.

1. Does the ball have an initial velocity?
2. Is this velocity increasing or decreasing?
3. Is the ball ever stationary and, if so, at what point?
4. If we ignore air resistance, what is the only force acting on the ball?
5. As the ball travels down, what is happening to its velocity?

What did you find?

1. As the ball leaves your hand travelling vertically upwards, it has a positive initial velocity.
2. This velocity is decreasing. As the ball gets higher and higher it slows down.
3. The ball’s velocity slows down until, for a moment, it is stationary. This is the balls maximum height.
4. Ignoring air resistance, the only force acting on the ball from the moment it leaves your hand is the force of acceleration due to gravity which, because it acts down is [latex]\scriptsize -9.8~\text{m.s}^{-2}[/latex].
5. As the ball falls, its velocity increases. However, the ball’s velocity is in the negative direction.

In activity 1.2 we dealt with an object (a ball in this case) that was thrown vertically up and fell vertically down and the only force the object experienced throughout was acceleration due to gravity. When objects travel vertically upwards and/or downwards while only experiencing acceleration due to gravity, we call this vertical projectile motion.

Activity 1.2 showed us a few important things about vertical projectile motion.

Take note!

• If we ignore air resistance, the only force acting on an object is acceleration due to gravity which is a constant [latex]\scriptsize g=9.8~\text{m.s}^{-2}[/latex].
• Objects moving upwards or downwards in the earth’s gravitational field, always accelerate downwards with this constant acceleration. Remember that non-zero acceleration means that the velocity is changing.
• Objects that have an initial velocity upwards will have zero velocity at their greatest height ([latex]\scriptsize h_{max}[/latex]). Remember that even though the velocity decreases to zero and then increases in the opposite direction, the acceleration experienced by the object is always constant.

Time symmetry

There is one aspect of free-falling bodies with initial upwards velocity that can be very helpful when solving problems and this is called time symmetry.

Consider an object thrown upwards from a vertical height, say [latex]\scriptsize h[/latex]. We know that it will travel upwards with decreasing velocity (there is a constant downwards acceleration) until it stops and starts to fall back down to the ground (increasing in velocity in the opposite direction).

The time that it takes for the object to fall back down to the same height [latex]\scriptsize h[/latex] is the same as the time taken for the object to reach its maximum height [latex]\scriptsize h_{max}[/latex] from height [latex]\scriptsize h[/latex] originally. This happens because the object experiences uniform acceleration throughout.

In addition, and again, because the object experiences a constant acceleration, the magnitude of the velocity at which the object reaches height [latex]\scriptsize h[/latex] on its way down is the same as the magnitude of the velocity with which the object was initially thrown upwards at height [latex]\scriptsize h[/latex]. Obviously, the velocities are opposite in sign.

Take note!

Time symmetry (shown in figure 3) is a consequence of the fact that a falling object experiences a constant acceleration ([latex]\scriptsize g=9.8~\text{m.s}^{-2}[/latex]).

• The time taken ([latex]\scriptsize t_1[/latex]) for the object to move from an initial height [latex]\scriptsize h[/latex] to [latex]\scriptsize h_{max}[/latex] is equal to the time taken ([latex]\scriptsize t_2[/latex]) for the object to move from [latex]\scriptsize h_{max}[/latex] back to the initial height [latex]\scriptsize h[/latex].
• The velocity ([latex]\scriptsize v_1[/latex]) of the object at height [latex]\scriptsize h[/latex] moving upwards is equal but opposite to the velocity ([latex]\scriptsize v_2[/latex]) of the object at height [latex]\scriptsize h[/latex] moving downwards.

Equations of motion of falling objects

The equations of motion that you learnt about and used in levels 2 and 3 are equally applicable to the motion of falling objects. Why? These equations apply to objects moving in a straight line and with a constant acceleration. Falling objects (objects moving up or down vertically) move in a straight line with a constant acceleration. In all these cases we know that the acceleration is due to gravity. Therefore, we know that [latex]\scriptsize a=g=9.8~\text{m.s}^{-2}[/latex].

We often refer to the movement of objects vertically up or down as vertical projectile motion.

Here is a reminder of the equations of motion you should know.

• [latex]\scriptsize v_f=v_i+at[/latex]
• [latex]\scriptsize s=v_it+\displaystyle \frac{1}{2}at^2[/latex]
• [latex]\scriptsize {v_f}^2={v_i}^2+2as[/latex]

where, in the case of vertical projectile motion:

• [latex]\scriptsize v_i[/latex] is the object’s initial velocity ([latex]\scriptsize \text{m.s}^{-1}[/latex])
• [latex]\scriptsize v_f[/latex] is the object’s initial velocity ([latex]\scriptsize \text{m.s}^{-1}[/latex])
• [latex]\scriptsize s[/latex] is the object’s vertical displacement ([latex]\scriptsize m[/latex]) or its change in vertical position ([latex]\scriptsize x_f-x_i[/latex])
• [latex]\scriptsize t[/latex] is the time interval ([latex]\scriptsize s[/latex])
• [latex]\scriptsize a[/latex] is the acceleration due to gravity ([latex]\scriptsize a=g=9.8~\text{m.s}^{-2}[/latex]).

Take note!

Remember that when we use the equations of motion, we are dealing with vectors which have magnitude and direction. Therefore, you need to decide which direction will be the positive direction so that your vectors have the correct signs.

Example 1.1

A ball is thrown vertically upwards with an initial velocity of [latex]\scriptsize 7~\text{m.s}^{-1}[/latex] from a height of [latex]\scriptsize 1~\text{m}[/latex]. Ignore air resistance.

1. What is the maximum height above the thrower’s hand reached by the ball?
2. What is the maximum height above the ground reached by the ball?
3. How long does it take the ball to reach its maximum height?
4. How long after reaching its maximum height will the ball hit the ground?
5. How long after leaving the thrower’s hand will the ball hit the ground?

Solutions

We are told that the ball is thrown vertically upwards and that we can ignore air resistance. Therefore, we are dealing with vertical projectile motion where the ball experiences a constant acceleration due to gravity. This is the only force acting on the ball.

1. We know that the ball’s velocity when it reaches its maximum height will be [latex]\scriptsize 0~\text{m.s}^{-1}[/latex]. Therefore, we know the ball’s initial and final velocity and we can calculate its displacement. We will assign upwards as the positive direction.
   [latex]\scriptsize \begin{align*} v_i&=7~\text{m.s}^{-1}\\ v_f&=0~\text{m.s}^{-1}\\ a&=g=-9.8~\text{m.s}^{-2}&&\text{Remember that upwards is}\\ &&&\text{positive and acceleration}\\ &&&\text{due to gravity always acts down}\\ {v_f}^2&={v_i}^2+2as\\ \therefore 2as&={v_f}^2-{v_i}^2\\ \therefore s&=\displaystyle \frac{{v_f}^2-{v_i}^2}{2a}\\ &=\displaystyle \frac{\left( 0~\text{m.s}^{-1} \right )^2- \left( 7~\text{m.s}^{-1} \right )^2}{2\times \left( -9.8~\text{m.s}^{-2} \right )}\\ &=\displaystyle \frac{-49~\text{m}^2. \text{s}^{-2}}{-19.6~\text{m.s}^{-2}}\\ &=2.5~\text{m} \end{align*}[/latex]
   .
   The ball’s maximum height above the thrower’s hand is [latex]\scriptsize 2.5~\text{m}[/latex].
2. The ball was initially thrown from a height of [latex]\scriptsize 1~\text{m}[/latex] above the ground. If its maximum height above the thrower’s hand is [latex]\scriptsize 2.5~\text{m}[/latex], then its maximum height above the ground is [latex]\scriptsize 2.5~\text{m}+1~\text{m}=3.5~\text{m}[/latex].
3. We know that the ball’s maximum height above the thrower’s hand is [latex]\scriptsize 2.5~\text{m}[/latex] and that its velocity at this height is [latex]\scriptsize 0~\text{m.s}^{-1}[/latex]. Therefore, we have two equations we can use to calculate time.
   [latex]\scriptsize v_f=v_i+at[/latex]
   [latex]\scriptsize s=v_it+\displaystyle \frac{1}{2}at^2[/latex]
   .
   It will be simpler to use the first of these:
   [latex]\scriptsize \begin{align*} v_i&=7~\text{m.s}^{-1}\\ v_f&=0~\text{m.s}^{-1}\\ a&=g=-9.8~\text{m.s}^{-2}&&\text{Remember that upwards is}\\ &&&\text{positive and acceleration}\\ &&&\text{due to gravity always acts down}\\ v_f&=v_i+at\\ \therefore at&=v_f-v_i\\ \therefore t&=\displaystyle \frac{v_f-v_i}{a}\\ &=\displaystyle \frac{ 0~\text{m.s}^{-1}-7~\text{m.s}^{-1}}{-9.8~\text{m.s}^{-2}}\\ &=0.71~\text{s} \end{align*}[/latex]
   .
   It will take [latex]\scriptsize 0.71~\text{s}[/latex] for the ball to reach its maximum height.
4. We know from question 2 that the ball’s maximum height above the ground is [latex]\scriptsize 3.5~\text{m}[/latex]. We need to calculate how long it will take the ball to travel this distance.
   [latex]\scriptsize \begin{align*} v_i&=0~\text{m.s}^{-1}\\ s&=-3.5~\text{m}&&\text{We need to be careful here.}\\ &&&\text{The change in the ball's}\\ &&&\text{vertical position is}\\ &&&x_f-x_i=0~\text{m}-3.5~\text{m}=-3.5~\text{m}\\ a&=g=-9.8~\text{m.s}^{-2}&&\text{Remember that upwards is}\\ &&&\text{positive and acceleration}\\ &&&\text{due to gravity always acts down}\\ s&=v_it+\displaystyle \frac{1}{2}at^2\\ \text{But }v_i&=0~\text{m.s}^{-1}\\ \therefore s&=\displaystyle \frac{1}{2}at^2\\ \therefore t^2&=\displaystyle \frac{2s}{a}\\ \therefore t&=\sqrt{\displaystyle \frac{2s}{a}}\\ &=\sqrt{\displaystyle \frac{2\times \left(-3.5~\text{m} \right )}{-9.8~\text{m.s}^{-2}}}\\ &=0.85~\text{s} \end{align*}[/latex]
   .
   It will take the ball [latex]\scriptsize 0.85~\text{s}[/latex] to reach the ground from its maximum height.
5. We know that the ball takes [latex]\scriptsize 0.71~\text{s}[/latex] to reach its maximum height and then another [latex]\scriptsize 0.85~\text{s}[/latex] to fall to the ground. Therefore, the total time is .[latex]\scriptsize 0.71~\text{s}+0.85~\text{s}=1.56~\text{s}[/latex].

In example 1.1 we could have assigned downwards as the positive direction. All our answers would have been the same. Why not redo the example and assign downwards as the positive direction to check?

Example 1.2

Question adapted from Everything Science Grade 12 Worked example 2 page 80

A cricketer hits a cricket ball so that it goes vertically upwards. If the ball takes to return to the initial height, determine its maximum height above the initial position.

Solution

In this question we will assign downwards as the positive direction.

We need to determine the maximum height of the ball. We are told that the ball takes [latex]\scriptsize 10~\text{s}[/latex] after leaving the bat to rise to the maximum height and then to fall back down to the initial height. We know, given time symmetry, that the ball will spend the same amount of time travelling up to this height as it does to fall back down to this height. Therefore, the ball will take [latex]\scriptsize 5~\text{s}[/latex] to get to its maximum height.

We also know, again given time symmetry, that its initial velocity on leaving the bat will be the same (but opposite) to its velocity as it falls back to the initial height.

Finally, we know that at its maximum height its velocity will be [latex]\scriptsize 0~\text{m.s}^{-1}[/latex].

If we focus on the portion of the ball’s flight from its maximum height back to the initial height, we can use the fact that [latex]\scriptsize v_i=0~\text{m.s}^{-1}[/latex].

[latex]\scriptsize \begin{align*} v_i&=0~\text{m.s}^{-1}\\ t&=5~\text{s}\\ a&=g=9.8~\text{m.s}^{-2}&&\text{Remember that downwards is}\\ &&&\text{positive and acceleration}\\ &&&\text{due to gravity always acts down}\\ s&=v_it+\displaystyle \frac{1}{2}at^2\\ \text{But }v_i&=0~\text{m.s}^{-1}\\ \therefore s&=\displaystyle \frac{1}{2}at^2\\ &=\displaystyle \frac{1}{2}\times 9.8~\text{m.s}^{-2}\times \left(5~\text{s} \right )^2\\ &=122.5~\text{m} \end{align*}[/latex]

Because the ball travels [latex]\scriptsize 122.5~\text{m}[/latex] to get back to its initial height from its maximum height, we know that the ball’s maximum height above the bat is [latex]\scriptsize 122.5~\text{m}[/latex].

Note: The positive answer for displacement in this case tells us that the ball’s change in position is downwards (it is falling) because we assigned downwards as positive.

It is important to note that in example 1.2 we did not calculate the ball’s maximum height above the ground, just its maximum height above the bat. We do not know how high the bat is from the ground, and hence do not know the ball’s maximum height above the ground.

This is important because it shows us that we can assign whatever coordinate system is convenient. In this case, we set our coordinate system to assign the zero position to the bat, not the ground.

Exercise 1.1

1. A soccer player kicks a soccer ball straight up into the air. The ball has an initial velocity of [latex]\scriptsize 15~\text{m.s}^{-1}[/latex].
   1. What height does the ball reach above the player’s boot before it stops to fall back to the ground?
   2. How long does it take for the ball to reach its maximum height?
2. A child throws a toy vertically upwards so that the toy reaches a maximum height of [latex]\scriptsize 2~\text{m}[/latex].
   1. What initial velocity did the toy have?
   2. How long does it take for the toy to reach its maximum height?
3. A tourist walking up the Leaning Tower of Pisa accidently drops her ice cream cone straight down. If the vertical distance covered in the last second is equal to the distance covered in first two seconds, find the height from which she dropped her ice-cream.
   

   

4. A scientist experimenting on a new mechanism of distributing food aid from airplanes drops a package from a hot-air balloon [latex]\scriptsize 60~\text{m}[/latex] off the ground. At the moment of the drop, the balloon is ascending vertically upwards with a velocity of [latex]\scriptsize 5~\text{m.s}^{-1}[/latex].
   1. With what velocity does the package hit the ground?
   2. How long does the package take to hit the ground?

The full solutions can be found at the end of the unit.

Summary

In this unit you have learnt the following:

• Vertical projectile motion deals with objects that fall vertically due to gravity alone.
• Objects falling due to gravity experience a constant acceleration of [latex]\scriptsize 9.8~\text{m.s}^{-2}[/latex] downwards.
• Because of this constant acceleration, objects fall with time symmetry, meaning that the time it takes an object to rise to its maximum height, is the same as the time it takes to fall back to its initial height.
• An object’s velocity at its maximum height is zero.
• The velocity with which an object that is thrown vertically upwards will be the same but opposite to the velocity the object will have when it falls back to this initial position.
• The equations of motion can be used to solve vertical projectile problems.

Unit 1: Assessment

Suggested time to complete: 15 minutes

1. Two balls, A and B, are simultaneously thrown into the air from the same height above the ground. Ball A is thrown upwards, and ball B is thrown downwards with the same speed.
   1. What can you say about the velocity of both balls as they hit the ground?
   2. What can you say about the acceleration both balls experience?
2. What acceleration will a ball thrown vertically in the air experience at its maximum height?

Question 3 adapted from NC(V) Level 4 Physical Science Paper 1 November 2015 question 6

3. An object is projected vertically upwards at [latex]\scriptsize 6~\text{m.s}^{-1}[/latex] from the roof of a building. It strikes a concrete surface below after with a velocity of [latex]\scriptsize 14~\text{m.s}^{-1}[/latex]. Ignore the effects of friction.
   
   1. How long will it take the object to reach its maximum height?
   2. Calculate the maximum height above the roof that the object reaches.
   3. Calculate the height of the building.

The full solutions can be found at the end of the unit.

Unit 1: Solutions

Exercise 1.1

1. .
   1. Let upwards be positive.
      [latex]\scriptsize \begin {align*} v_i &= 15~\text{m.s}^{-1}\\ v_f &= 0~\text{m.s}^{-1}\\ a &= -9.8~\text{m.s}^{-2}\\ {v_f}^{2} &= {v_i}^{2}+2as\\ \therefore s&=\displaystyle \frac{{v_f}^{2}–{v_i}^{2}}{2a}\\ &=11.48~\text{m} \end {align*}[/latex]
   2. .
      [latex]\scriptsize \begin {align*} v_i &= 15~\text{m.s}^{-1}\\ v_f &= 0~\text{m.s}^{-1}\\ a &= -9.8~\text{m.s}^{-2}\\ {v_f} &= {v_i}+at\\ \therefore t&=\displaystyle \frac{{v_f}–{v_i}}{a}\\ &=1.53~\text{s} \end {align*}[/latex]
2. .
   1. Let upwards be positive
      [latex]\scriptsize \begin {align*} s &= 2~\text{m}\\ v_f &= 0~\text{m.s}^{-1}\\ a &= -9.8~\text{m.s}^{-2}\\ {v_f}^{2} &= {v_i}^{2}+2as\\ \therefore {v_i}^{2}&={v_f}^{2}–2as\\ \therefore v_i&=\sqrt{{v_f}^{2}–2as}\\ &=6.26~\text{m.s}^{-1} \end {align*}[/latex]
      .
      The ball had an initial velocity of [latex]\scriptsize 6.26~\text{m.s}^{-1}[/latex] upwards.
   2. .
      [latex]\scriptsize \begin {align*} v_i&=6.26~\text{m.s}^{-1}\\ v_f&=0~\text{m.s}^{-1}\\ a&=-9.8~\text{m.s}^{-2}\\ v_f&=v_i+at\\ \therefore t&=\displaystyle \frac{v_f-v_i}{a}\\ &=0.64~\text{s} \end{align*}[/latex]
3. Let upwards be positive.
   First two seconds:
   [latex]\scriptsize \begin {align*} v_i&=0~\text{m.s}^{-1}\\ t&=2~\text{s}\\ s&=v_i t+\displaystyle \frac{1}{2}at^2\\ &=\displaystyle \frac{1}{2}\times 4\times a\\ &=2a \end{align*}[/latex]
   .
   Last second:
   [latex]\scriptsize \begin {align*} t&=1~\text{s}\\ s&=v_i t+\displaystyle \frac{1}{2}at^2\\ &=v_i+\displaystyle \frac{1}{2}a\\ \text{But }s&=2a\\ \therefore 2a&=v_i+\displaystyle \frac{1}{2}a\\ \therefore v_i&=\displaystyle \frac{3}{2}a \end{align*}[/latex]
   .
   Last second:
   [latex]\scriptsize \begin {align*} v_f&=v_i+at\\ &=\displaystyle \frac{3}{2}a+a\\ &=\displaystyle \frac{5}{2}a \end{align*}[/latex]
   .
   We know that for the full fall:
   [latex]\scriptsize \begin {align*} v_i&=0~\text{m.s}^{-1}\\ v_f&=\displaystyle \frac{5}{2}a\\ a&=-9.8~\text{m.s}^{-1}\\ {f_v}^{2}&={v_i}^{2}+2as\\ \therefore s&=\displaystyle \frac{{f_v}^{2}-{v_i}^{2}}{2a}\\ &=\displaystyle \frac{\displaystyle \frac{25}{4}a^{2}}{2a}\\ &=\displaystyle \frac{25}{8}a\\ &=-30.625~\text{s} \end{align*}[/latex]
   .
   Therefore, the ice-cream was dropped from a height of [latex]\scriptsize 30.625~\text{m}[/latex].
4. .
   1. Let upwards be positive.
      To calculate the velocity with which the package hits the ground, we need to calculate the maximum height above the ground from which the package falls.
      .
      [latex]\scriptsize \begin {align*} v_i &= 5~\text{m.s}^{-1}\\ v_f &= 0~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ {v_f}^2 &= {v_i}^2 + 2as\\ \therefore s&= \displaystyle \frac{{v_f}^2-{v_i}^2}{2a}\\ &= 1.28~\text{m} \end{align*}[/latex]
      .
      Therefore, the maximum height of the package above the ground is [latex]\scriptsize 60~\text{m} + 1.28~\text{m} = 61.28~\text{m}[/latex].
      .
      [latex]\scriptsize \begin {align*} v_i &= 0~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ s&=-61.28~\text{m}\\ {v_f}^2 &={v_i}^2 + 2as\\ \therefore v_f&=\sqrt{{v_i}^2 + 2as}\\ &=-34.66~\text{m.s}^{-1} \end{align*}[/latex]
      .
      Therefore the package will have a velocity of [latex]\scriptsize 34.66~\text{m.s}^{-1}[/latex] downwards when it hits the ground.
   2. .
      [latex]\scriptsize \begin {align*} v_i &= 0~\text{m.s}^{-1}\\ v_f&=-34.66~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ v_f &=v_i + at\\ \therefore t&=\displaystyle \frac{v_f-v_i}{a}\\ &=3.54~\text{s} \end{align*}[/latex]

Back to Exercise 1.1

Unit 1: Assessment

1. .
   1. The velocity of both balls will be the same as they hit the ground. By the time ball A returns to the point from which it was thrown, its velocity will have the same magnitude with which it was initially thrown but in the downward direction. This will be the same velocity with which ball B was thrown.
   2. Both balls experience the same constant acceleration due to gravity [latex]\scriptsize – 9.8 ~\text{m.s}^{-2}[/latex] downwards.
2. [latex]\scriptsize 9.8 ~\text{m.s}^{-2}[/latex] downwards
3. .
   1. Let upwards be positive.
      [latex]\scriptsize \begin {align*} v_i &= 6~\text{m.s}^{-1}\\ v_f&=0~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ v_f &=v_i + at\\ \therefore t&=\displaystyle \frac{v_f-v_i}{a}\\ &=0.61~\text{s} \end{align*}[/latex]
   2. .
      [latex]\scriptsize \begin {align*} v_i &= 6~\text{m.s}^{-1}\\ v_f&=0~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ t&=0.61~\text{s}\\ s&=v_it+\displaystyle \frac{1}{2}at^2\\ &=1.84~\text{m} \end{align*}[/latex]
      .
      The ball will reach a maximum height above the roof of [latex]\scriptsize 1.84~\text{m}[/latex].
   3. The ball strikes the ground [latex]\scriptsize 5~\text{s}[/latex] after being thrown. Therefore, it strikes the ground [latex]\scriptsize 5~\text{s} – 0.61~\text{s}=4.39~\text{s}[/latex] after reaching its maximum height. Its velocity on hitting the ground is [latex]\scriptsize 14~\text{m.s}^{-1}[/latex].
      [latex]\scriptsize \begin {align*} v_i &= 0~\text{m.s}^{-1}\\ v_f&=-14~\text{m.s}^{-1}\\ a &= -9.8 \text{m.s}^{-2}\\ t&=4.93~\text{s}\\ {v_f}^2&={v_i}^2+2as\\ \therefore s&=\displaystyle \frac{{v_f}^2-{v_i}^2}{2a}\\ &=-10~\text{m} \end{align*}[/latex]
      .
      Therefore, the ball fell [latex]\scriptsize 10~\text{m}[/latex] from its maximum height. But, its maximum height above the building was [latex]\scriptsize 1.84~\text{m}[/latex], therefore, the building is [latex]\scriptsize 10~\text{m} – 1.84~\text{m} = 8.16~\text{m}[/latex] high.

Back to Unit 1: Assessment