Interactive textbook: Projectiles

Motion under gravity

When a ball is thrown vertically it travels upwards while slowing down, then stops momentarily and then comes back down. If we take 'up' to be positive, the acceleration of the ball is negative both on the way up and on the way down. By measuring the distance travelled and time we can deduce that the acceleration due to gravity has a constant value close to the surface of the Earth: \(g=-9.81\text{ m s}^{-2}\)

Measuring acceleration due to gravity

One way to measure the acceleration due to gravity is to analyse a video of a falling ball using a programme such as LoggerPro.

The position of the ball is marked at regular time intervals and a graph of displacement vs time is plotted:

The best-fit line is a parabola showing that the displacement is proportional to \(t^2\), which means that acceleration is uniform. The \(A\) value of the quadratic is equal to \(g\over 2\) so \(g=-9.8\text{ m s}^{-2}\).

From these data, LoggerPro can calculate the velocity for each time interval. A graph of velocity vs time can then be plotted:

A similar experiment could be performed for a body being thrown upwards:

This shows how the acceleration is the same on the way up as it is on the way down, \(g=-9.8\text{ m s}^{-2}\).

 

This is a displacement time graph for a body thrown vertcally upwards

What is the intial velocity?

Initial gradient = 8 m s-1

Projectile motion

If a ball is thrown forwards it will have two components of velocity:

Horizontal: constant velocity

Vertical: constant acceleration

This leads to a parabolic path as can be seen in the animation below. Watch this a few times considering the horizontal and then vertical motion:

The blue dot has constant horizontal velocity of \(1\text{ m s}^{-1}\) and the red dot has vertical motion with constant negative acceleration of \(-2\text{ m s}^{-2}\) and initial velocity \(1\text{ m s}^{-1}\). The black dot combines these two components of motion with a parabolic trajectory.

The trajectory is the same as the graph of displacement vs time for the horizontal motion. This is because the horizontal velocity is \(1\text{ m s}^{-1}\) so the dot travels \(1\text{ m}\) in every second.

Let's analyse the motion:

Since the two components are perpendicular to each other they can be treated independently.

Vertical components

Vertically, the ball has constant acceleration so we can use the 'suvat' equations. For the complete flight:

  • \(s = 0\) (the dot starts and ends at the same height)
  • \(u = 1\text{ m s}^{-1}\)
  • \(v = -1\text{ m s}^{-1}\) (the motion is symmetrical)
  • \(a = -2\text{ m s}^{-2}\)

The only unknown is the time of flight, we can find this in two ways:

\(a={v-u\over t} \Rightarrow t={v-u\over a}={-1-1\over -2}=1\text{ s}\)

\(s=ut+{1\over 2}at^2\Rightarrow 0=1t+{1\over 2}(-2)t^2\Rightarrow t=1\text{ s}\)

The maximum height reached is the vertical displacement after half the time of flight, 0.5 s:

\(s=ut+{1\over 2}at^2=1\times0.5+{1\over2}(-2)\times0.5^2=0.25\text{ m}\)

Alternaitvely the maximum height can be determined using the fact that the vertical component of velocity at the top is zero:

\(s={1\over2}(v+u)t={1\over2}(0+1)\times 0.5=0.25\text{ m}\)

Horizontal components

Horizontally, the ball has constant velocity so the only equation we require is:

\(s_{H}=v_\text{H}t\)

The total time of flight is \(1\text{ s}\) so the horiztonal range can be calculated as:

\(s_\text{H}=1\times 1=1\text{ m}\)

Note that all projectiles thrown on Earth have a vertical acceleration of \(-9.8\text{ m s}^{-2}\).

A firework is propelled by the burning chemicals coming out of it until they run out at a height of 100m.

The horizontal component of velocity  from the ground to 100 m and from 100 m back to the ground is:

The chemicals cause acceleration in both components of motion.

Components of velocity

In typical projectile motion problems you won't be given the vertical and horizontal components of velocity. Instead, you will given the speed and angle to the horizontal and have to find the components yourself.

Let's consider a ball kicked at an angle of \(30°\) from the ground at \(20\text{ m s}^{-1}\):

The vertical component of the initial velocity is \(20 \sin30° = 10\text{ m s}^{-1}\).The horizontal component of the initial velocity (which remains constant throughout) is \(20 \cos30° = 17.3\text{ m s}^{-1}\).

The time of flight can be found using the vertical component (here taking \(g = -10\text{ m s}^{-2}\)):

\(s=ut+{1\over2}at^2\)

\(0=10t-5t^2\Rightarrow t=2\text{ s}\)

Using half the time we can find the maximum height:

\(s={1\over 2}(v+u)t={1\over2}(10+0)\times 1=5\text{ m}\)

The range can be found using the horizontal component of velocity and the total time:

\(s_\text{H}=v_\text{H}t=17.3\times 2=34.6\text{ m}\)

Note that the maximum height is a little high since it is measured from the centre of the ball.

 

A ball is thrown with a velocity of 5 m s-1 at an angle of 60° to the ground.

What is the speed of the ball when it reaches it's maximum height?

At the top the ball has zero vertical velocity so the speed = horizontal velocity which is the same as the intial horizontal component.

5 x cos60

Components of velocity can also added together using Pythagoras (to give magnitude) and trigonometry (to give direction). This could be useful if calculating the overall velocity with which a stone kicked from a cliff strikes the sea.

Projectiles with air resistance

If you were to kick a ball as in the previous example it wouldn't go so far or so high because air resistance opposes the motion. Horizontally, this has the effect of introducing a non-uniform negative acceleration. Vertically it increases the magnitude of acceleration on the way up and reduces it on the way down. This is too complicated to deal with mathematically but can be shown in the simulation.

The first ball is without air resistance and the the second is with. The cloud is a cloud   

A golf ball experiences an upward force due the the flow of air around it.

What effect does this have on the range compared to the ball travelling in a vacuum.

The upward force reduces the vertical acceleration of the ball causing the time of flight to be longer.

 

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