Physics A Level
Chapter 2: Accelerated motion 2.14 Understanding projectiles
Physics A Level
Chapter 2: Accelerated motion 2.14 Understanding projectiles
We will first consider the simple case of a projectile thrown straight up in the air, so that it moves vertically. Then we will look at projectiles that move horizontally and vertically at the same time.
Up and down
A stone is thrown upwards with an initial velocity of $20\,m\,{s^{ - 1}}$. Figure 2.30 shows the situation.
It is important to use a consistent sign convention here. We will take upwards as positive, and downwards as negative. So the stone’s initial velocity is positive, but its acceleration g is negative. We can solve various problems about the stone’s motion by using the equations of motion.
How high?
How high will the stone rise above ground level of the cliff?
As the stone rises upwards, it moves more and more slowly – it decelerates because of the force of gravity.
At its highest point, the stone’s velocity is zero. So the quantities we know are:
$\begin{array}{l}
initial\,velocity\, = \,u\, = \,20\,m\,{s^{ - 1}}\\
final\,velocity\, = \,v\, = \,0\,m\,{s^{ - 1}}\\
acceleration\, = \,a\, = \, - 9.81\,m\,{s^{ - 1}}\\
displacement\, = \,s\, = \,?
\end{array}$
The relevant equation of motion is ${v^2} = {u^2} + 2as$. Substituting values gives:
$\begin{array}{l}
{0^2} = {20^2} + 2 \times ( - 9.81) \times s\\
{0^2} = 400 - 19.62\,s\\
s = \frac{{400}}{{19.62}}\\
= 20.4\,m\, \approx \,20\,m
\end{array}$
The stone rises $20 m$ upwards before it starts to fall again.
How long?
How long will it take from leaving your hand for the stone to fall back to the clifftop?
When the stone returns to the point from which it was thrown, its displacement s is zero. So:
$\begin{array}{l}
s = 0\\
u = 20\,m\,{s^{ - 1}}\\
a = - 9.81\,m\,{s^{ - 2}}\\
t = ?
\end{array}$
Substituting in $s = ut + \frac{1}{2}a{t^2}$ gives:
$\begin{array}{l}
0 = 20t \times \frac{1}{2}( - 9.81) \times {t^2}\\
= 20t - 4.905{t^2}\\
= (20 - 4.905t) \times t
\end{array}$
There are two possible solutions to this:
- $t = 0s$; in other words, the stone had zero displacement at the instant it was thrown
- $t = 4.1 s$; in other words, the stone returned to zero displacement after $4.1 s$, which is the answer we are interested in.
Falling further
The height of the cliff is $25 m$. How long will it take the stone to reach the foot of the cliff?
This is similar to the last example, but now the stone’s final displacement is $25 m$ below its starting point.
By our sign convention, this is a negative displacement and $s = −25 m$.
Questions
22) In the example in ‘Falling further’, calculate the time it will take for the stone to reach the foot of the cliff.
23) A ball is fired upwards with an initial velocity of $30\,m\,{s^{ - 1}}$. Table 2.6 shows how the ball’s velocity changes. (Take $g = 9.81\,m\,{s^{ - 2}}$.)
a: Copy and complete the table.
b: Draw a graph to represent the data.
c: Use your graph to deduce how long the ball took to reach its highest point.
| 20.19 | 30 | $Velocity\,/\,m\,{s^{ - 1}}$ | ||||
| 5.0 | 4.0 | 3.0 | 2.0 | 1.0 | 0 | Time/s |
Vertical and horizontal at the same time
Here is an example to illustrate what happens when an object travels vertically and horizontally at the same time.
In a toy, a ball-bearing is fired horizontally from a point $0.4 m$ above the ground. Its initial velocity is $2.5\,m\,{s^{ - 1}}$. Its positions at equal intervals of time have been calculated and are shown in Table 2.7. These results are also shown in Figure 2.31. Study the table and the graph. You should notice the following:
- The horizontal distance increases steadily. This is because the ball’s horizontal motion is unaffected by the force of gravity. It travels at a steady velocity horizontally so we can use $v = \frac{s}{t}$.
- The vertical distances do not show the same pattern. The ball is accelerating downwards so we must use the equations of motion. (These figures have been calculated using $g = 9.81\,m\,{s^{ - 2}}$.)
| Vertical distance / m | Horizontal distance / m | Time / s |
| 0.000 | 0.00 | 0.00 |
| 0.008 | 0.10 | 0.04 |
| 0.031 | 0.20 | 0.08 |
| 0.071 | 0.30 | 0.12 |
| 0.126 | 0.40 | 0.16 |
| 0.196 | 0.50 | 0.20 |
| 0.283 | 0.60 | 0.24 |
| 0.385 | 0.70 | 0.28 |
horizontal and vertical components of its velocity
You can calculate the distance s fallen using the equation of motion $s = ut + \frac{1}{2}a{t^2}$. (The initial vertical velocity $u = 0$.)
The horizontal distance is calculated using:
horizontal distance $ = 2.5 \times t$
The vertical distance is calculated using:
$vertical\,dis\tan ce = \frac{1}{2} \times 9.81 \times {t^2}$
WORKED EXAMPLES
9) A stone is thrown horizontally with a velocity of $12\,m\,{s^{ - 1}}$ from the top of a vertical cliff.
Calculate how long the stone takes to reach the ground $40 m$ below and how far the stone lands from the base of the cliff.
Step 1: Consider the ball’s vertical motion. It has zero initial speed vertically and travels $40 m$ with acceleration $9.8\,m\,{s^{ - 2}}$ in the same direction.
$\begin{array}{l}
s = ut + \frac{1}{2}a{t^2}\\
40 = 0 + \frac{1}{2} \times 9/81 \times {t^2}
\end{array}$
So, $t = 2.86 s$.
Step 2: Consider the ball’s horizontal motion. The ball travels with a constant horizontal velocity, $12\,m\,{s^{ - 1}}$, as long as there is no air resistance.
distance travelled $ = u \times t = 12 \times 2.86 = 34.3\,m$
Hint: You may find it easier to summarise the information like this:
vertically $s = 40\,u = 0\,a = 9.81\,t = ?\,v = ?$
horizontally $u = 12\,v = 12\,a = 0\,t = ?\,s = ?$
10) A ball is thrown with an initial velocity of $20\,m\,{s^{ - 1}}$ at an angle of ${30^ \circ }$ to the horizontal (Figure 2.32).
Calculate the horizontal distance travelled by the ball (its range).
Step 1: Split the ball’s initial velocity into horizontal and vertical components:
initial velocity $ = u = 20\,m\,{s^{ - 1}}$
horizontal component of initial velocity $v = u\cos \theta = 20 \times \cos \,{30^ \circ } = 17.3\,m\,{s^{ - 1}}$
vertical component of initial velocity $v = u\sin \theta = 20 \times \sin {30^ \circ } = 10\,m\,{s^{ - 1}}$
Step 2: Consider the ball’s vertical motion. How long will it take to return to the ground? In other words, when will its displacement return to zero?
$u = 10\,m\,{s^{ - 1}}\,\,\,a = - 9.81\,m\,{s^{ - 2}}\,\,\,\,s = 0\,\,\,\,t = ?$
Using $s = ut + \frac{1}{2}a{t^2}$ , we have:
$0 = 10t - 4.905{t^2}$
This gives $t = 0 s$ or $t = 2.04 s$.
So, the ball is in the air for $2.04 s$.
Step 3: Consider the ball’s horizontal motion. How far will it travel horizontally in the $2.04 s$ before it lands? This is simple to calculate, since it moves with a constant horizontal velocity of $17\,m\,{s^{ - 1}}$.
horizontal displacement $\begin{array}{l}
s = 17.3 \times 2.04\\
= 35.3m
\end{array}$
Hence the horizontal distance travelled by the ball (its range) is about $35 m$.
Questions
24) A stone is thrown horizontally from the top of a vertical cliff and lands $4.0 s$ later at a distance $12.0 m$ from the base of the cliff. Ignore air resistance.
a: Calculate the horizontal speed of the stone.
b: Calculate the height of the cliff.
25) A stone is thrown with a velocity of $8.0\,m\,{s^{ - 1}}$ into the air at an angle of ${40^ \circ }$ to the horizontal.
a: Calculate the vertical component of the velocity.
b: State the value of the vertical component of the velocity when the stone reaches its highest point.
Ignore air resistance.
c: Use your answers to part a and part b to calculate the time the stone takes to reach its highest point.
d: Calculate the horizontal component of the velocity.
e: Use your answers to part c and part d to find the horizontal distance travelled by the stone as it climbs to its highest point.
26) The range of a projectile is the horizontal distance it travels before it reaches the ground. The greatest range is achieved if the projectile is thrown at ${45^ \circ }$ to the horizontal.
A ball is thrown with an initial velocity of $40\,m\,{s^{ - 1}}$. Calculate its greatest possible range when air resistance is considered to be negligible.
EXAM-STYLE QUESTIONS
1) An aircraft, starting from rest accelerates uniformly along a straight runway. It reaches a speed of $200\,km\,{h^{ - 1}}$ and travels a distance of $1.4 km$.
What is the acceleration of the aircraft along the runway? [1]
A: $1.1\,m\,{s^{ - 2}}$
B: $2.2\,m\,{s^{ - 2}}$
C: $3.0\,m\,{s^{ - 2}}$
D: $6.0\,m\,{s^{ - 2}}$
2) A ball is thrown with a velocity of $10\,m\,{s^{ - 1}}$ at an angle of ${30^ \circ }$ to the horizontal.
Air resistance has a negligible effect on the motion of the ball.
What is the velocity of the ball at the highest point in its path? [1]
A: 0
B: $5.0\,m\,{s^{ - 1}}$
C: $8.7\,m\,{s^{ - 1}}$
D: $10\,m\,{s^{ - 1}}$
3) A trolley travels along a straight track. The variation with time t of the velocity v of the trolley is shown. [1]
Which graph shows the variation with time of the acceleration a of the trolley?
4) A motorway designer can assume that cars approaching a motorway enter a slip road with a velocity of $10\,m\,{s^{ - 1}}$ and reach a velocity of $30\,m\,{s^{ - 1}}$ before joining the motorway. Calculate the minimum length for the slip road, assuming that vehicles have an acceleration of $4.0\,m\,{s^{ - 2}}$. [4]
5) A train is travelling at $50\,m\,{s^{ - 1}}$ when the driver applies the brakes and gives the train a constant deceleration of magnitude $0.50\,m\,{s^{ - 2}}$ for $100 s$. Describe what happens to the train. Calculate the distance travelled by the train in $100s$. [7]
6) A boy stands on a cliff edge and throws a stone vertically upwards at time $t =0$. The stone leaves his hand at $20\,m\,{s^{ - 1}}$. Take the acceleration of the ball as $9.8\,m\,{s^{ - 2}}$.
a: Show that the equation for the displacement of the ball is:
$s = 20t - 4.9{t^2}$. [2]
b: Calculate the height of the stone $2.0 s$ after release and $6.0 s$ after release. [3]
c: Calculate the time taken for the stone return to the level of the boy’s hand.
You may assume the boy’s hand does not move vertically after the ball is released. [4]
[Total: 9]
7) This graph shows the variation of velocity with time of two cars, A and B, which are travelling in the same direction over a period of time of $40 s$.
Car A, travelling at a constant velocity of $40\,m\,{s^{ - 1}}$, overtakes car B at time $t =0$. In order to catch up with car A, car B immediately accelerates uniformly for $20 s$ to reach a constant velocity of $50\,m\,{s^{ - 1}}$. Calculate:
a: the distance that A travels during the first $20 s$. [2]
b: the acceleration and distance of travel of B during the first $20 s$ [5]
c: the additional time taken for B to catch up with A [2]
d: the distance each car will have then travelled since $t = 0$. [2]
[Total: 11]
8) An athlete competing in the long jump leaves the ground with a velocity of $5.6\,m\,{s^{ - 1}}$ at an angle of ${30^ \circ }$ to the horizontal.
a: Determine the vertical component of the velocity and use this value to find the time between leaving the ground and landing. [4]
b: Determine the horizontal component of the velocity and use this value to find the horizontal distance travelled. [4]
[Total: 8]
9) This diagram shows an arrangement used to measure the acceleration of a metal plate as it falls vertically.
The metal plate is released from rest and falls a distance of $0.200 m$ before breaking light beam 1. It then falls a further $0.250 m$ before breaking light beam 2.
a: Calculate the time taken for the plate to fall $0.200 m$ from rest. (You may assume that the metal plate falls with an acceleration equal to the acceleration of free fall.) [2]
b: The timer measures the speed of the metal plate as it falls through each light beam. The speed as it falls through light beam 1 is $1.92\,m\,{s^{ - 1}}$ and the speed as it falls through light beam 2 is $2.91\,m\,{s^{ - 1}}$.
i- Calculate the acceleration of the plate between the two light beams. [2]
ii- State and explain one reason why the acceleration of the plate is not equal to the acceleration of free fall. [2]
[Total: 6]
10) This is a velocity–time graph for a vertically bouncing ball.
The ball is released at A and strikes the ground at B. The ball leaves the ground at D and reaches its maximum height at E. The effects of air resistance can be neglected.
a: State:
i- why the velocity at D is negative [1]
ii- why the gradient of the line AB is the same as the gradient of line DE [1]
iii- what is represented by the area between the line AB and the time axis [1]
iv- why the area of triangle ABC is greater than the area of triangle CDE. [1]
b: The ball is dropped from rest from an initial height of $1.2 m$. After hitting the ground the ball rebounds to a height of $0.80 m$. The ball is in contact with the ground between B and D for a time of $0.16 s$.
Using the acceleration of free fall, calculate:
i- the speed of the ball immediately before hitting the ground [2]
ii- the speed of the ball immediately after hitting the ground [2]
iii- the acceleration of the ball while it is in contact with the ground. State the direction of this acceleration. [3]
[Total: 11]
11) A student measures the speed v of a trolley as it moves down a slope. The variation of v with time t is shown in this graph.
a: Use the graph to find the acceleration of the trolley when $t = 0.70 s$. [2]
b: State how the acceleration of the trolley varies between $t = 0$ and $t = 1.0 s$.
Explain your answer by reference to the graph. [3]
c: Determine the distance travelled by the trolley between $t = 0.60$ and $t =0.80 s$. [3]
d: The student obtained the readings for v using a motion sensor. The readings may have random errors and systematic errors. Explain how these two types of error affect the velocity–time graph. [2]
[Total: 10]
12) A car driver is travelling at speed v on a straight road. He comes over the top of a hill to find a fallen tree on the road ahead. He immediately brakes hard but travels a distance of $60 m$ at speed v before the brakes are applied. The skid marks left on the road by the wheels of the car are of length $140 m$, as shown.
The police investigate whether the driver was speeding and establish that the car decelerates at $2.0\,m\,{s^{ - 2}}$ during the skid.
a: Determine the initial speed v of the car before the brakes are applied. [2]
b: Determine the time taken between the driver coming over the top of the hill and applying the brakes. Suggest whether this shows whether the driver was alert to the danger. [2]
c: The speed limit on the road is $100\,km/h$. Determine whether the driver was breaking the speed limit. [2]
[Total: 6]
13) A hot-air balloon rises vertically. At time $t = 0$, a ball is released from the balloon. This graph shows the variation of the ball’s velocity v with t. The ball hits the ground at $t = 4.1 s$.
a: Explain how the graph shows that the acceleration of the ball is constant. [1]
b: Use the graph to:
i- determine the time at which the ball reaches its highest point [1]
ii- show that the ball rises for a further $12 m$ between release and its highest point [2]
iii- determine the distance between the highest point reached by the ball and the ground. [2]
c: The equation relating v and t is $v = 15 - 9.81t$. State the significance in the equation of:
i- the number 15 [1]
ii- the negative sign. [1]
[Total: 8]
14) An aeroplane is travelling horizontally at a speed of $80\,m\,{s^{ - 1}}$ and drops a crate of emergency supplies.
To avoid damage, the maximum vertical speed of the crate on landing is $20\,m\,{s^{ - 1}}$. You may assume air resistance is negligible.
a: Calculate the maximum height of the aeroplane when the crate is dropped. [2]
b: Calculate the time taken for the crate to reach the ground from this height. [2]
c: The aeroplane is travelling at the maximum permitted height. Calculate the horizontal distance travelled by the crate after it is released from the aeroplane. [1]
[Total: 5]
SELF-EVALUATION CHECKLIST
After studying the chapter, complete a table like this:
| I can | See topic | Needs more work | Almost there | Ready to move on |
| define acceleration | 2.1 | |||
| calculate displacement from the area under a velocity–time graph | 2.5 | |||
| calculate velocity using the gradient of a displacement–time graph | 2.6 | |||
| calculate acceleration using the gradient of a velocity–time graph | 2.4 | |||
| derive and use the equations of uniformly accelerated motion | 2.10 | |||
| describe an experiment to measure the acceleration of free fall, g | 2.11, 2.12 | |||
| use perpendicular components to represent a vector | 2.13 | |||
| explain projectile motion using uniform velocity in one direction and uniform acceleration in a perpendicular direction and do calculations on this motion. | 2.14 |