9231M. Motion of a Projectile

Section 1: Finding the horizontal distance travelled by particles

We consider projectiles moving under the force due to gravity. We consider the horizontal and vertical components of their displacement, velocity and acceleration using the SUVAT formulae (as if only gravity is acting on them they have constant acceleration).

In our models we typically:

  • ignore air resistance;
  • treat projectiles as particles;
  • ignore any rotational forces; and
  • assume that the force due to gravity is constant.

A projectile travels on a parabolic trajectory.

We already know that the displacement of a projectile is always s = ut + \frac{1}{2} at^2 , which is a quadratic function (graphically represented by a parabola).

If we imagine a particle projected at angle 𝜽 (anticlockwise from horizontal) with velocity u, we can resolve its initial velocity into horizontal and vertical directions as ucos𝜽 and usin𝜽.

So, using the SUVAT equation v = u + at, as the only acceleration acting on the particle is downwards due to gravity, we have horizontal velocity of vX=ucos𝜽 and vertical velocity of vY=usin𝜽 – gt.

At the vertex of the particle’s parabolic path, i.e. at its maximum height, vY=0, so from the above, t = \frac{u sin \theta}{g}

If the particle is projected from the ground, from symmetry we can see that the time it takes to travel to the top of its path is equal to the time it takes to return to the ground, so we can conclude that the total flight time is t = \frac{2u sin \theta}{g}.

Considering displacement, we have s_X = ucos \theta t and s_Y = usin \theta t + \frac{1}{2} at^2

If we substitute the time of the flight path from projection to returning to the ground into sX above, we get s_X = ucos \theta \times \frac{2u sin \theta}{g} , or s_X = \frac{u^2cos \theta sin \theta}{g} . We can use trigonometric double angle formulae to write this as s_X = \frac{u^2 sin 2 \theta}{g} . This is the range of the the projectile, its maximum horizontal difference.

We can see that when 𝜽 = 45°, so sin2𝜽 = 1 and the range is maximised. So an angle of elevation of 45° achieves maximum displacement.

Worked Example 1

A particle is projected from a point on a horizontal surface with initial speed 20ms-1 at an angle of elevation of 25°. Find:

  • the range of the particle;
  • the time taken to reach the highest point; and
  • the speed of the particle when t = 0.5.

A particle may be projected from a point above the ground (like when a tennis ball is hit). If we wish to find the velocity when the particle hits the ground, we can simply use the SUVAT formula v2 = u2+2as in the vertical direction, with the s indicating how far above the ground it is projected from and the above formula vX=ucos𝜽 in the horizontal direction, using Pytahgoras theorem with these two results to find the velocity when it hits the ground.

Worked Example 2

A particle is projected from the top of a platform 15m above the horizontal floor below. The angle of projection is 45° and the initial speed is 25ms-1. Find:

  • the exact time taken to reach the horizontal floor below; and
  • the exact speed of the particle as it hits the floor.

Projectiles can also be launched at an angle below the horizontal (an angle of depression), i.e pointing partly downwards. If we wish to find how far such a particle travels horizontally, we first must find how long it takes to reach the ground and then use this time value to calculate its horizontal displacement.

We solve the quadratic equation s = ut + \frac{1}{2} at^2 downwards, taking acceleration as positive and then substitute the positive result for t into s_X = utcos \theta

Worked Example 3

An aircraft is flying over the ocean to collect data. It is travelling horizontally at a speed of 50ms-1 when it launches a sensor at an angle of 25° below the horizontal with an initial speed of 10ms-1. Given that the plane is 1km above the ocean, and assuming that the ocean surface is flat and that air resistance is negligible, find the speed of the sensor when it is 100m from the surface of the ocean.

Exercise 1


  1. (a) 45.62ms-1 , (b) 88.68m, (c) 27.17° or 62.83°
  2. 42.75ms-1
  3. 14.59ms-1< u <16.31ms-1
  4. 36.9°
  5. 10√2 < u < 20
  6. 7.75ms-1
  7. (a) \sqrt{60} ms^{-1} (b) 2.75m
  8. 6 \sqrt{15} m
  9. 2.62s
  10. (a) 2.32s, (b) 25ms-1

Section 2: Cartesian equation of trajectory

We saw above that a particle’s initial speed and angle of projection determines its general motion. We found that the horizontal distance travelled at time t is x = utcos \theta , which can be rewritten as t = \frac{x}{ucos \theta}

We also found that the vertical distance travelled at time t is y = utsin \theta  - \frac{1}{2} gt^2 . If we substitute the above expression for t into this, we get: y = u(\frac{x}{ucos \theta})sin \theta  - \frac{1}{2} g(\frac{x}{ucos \theta})^2 , so, our Cartesian equation of the trajectory of a projectile is: y = xtan \theta - \frac{gx^2}{2u^2}sec^2 \theta

This is a quadratic equation, with a negative coefficient for x2, so represents a parabola opening downwards.

We can use this equation to find horizontal distance travelled (spoiler alert: it gives the same result as we got in the section above:

When a projectile returns to the ground, y = 0, so 0 = xtan \theta - \frac{gx^2}{2u^2}sec^2 \theta , so x(tan \theta - \frac{g}{2u^2}xsec^2 \theta)=0 , so the start and finish points of the motion are x=0 and x = \frac{u^2sin2 \theta}{g}

Worked Example 1

A projectile is launched from a point O, 1m above the horizontal ground level. It has initial speed of 20ms-1 and angle of elevation of 25°. Find the horizontal distance from O when the particle is 2m above the ground and descending.

If we consider downward projection, the only difference is that gravity is now effectively positive, so we have y = xtan \theta + \frac{gx^2}{2u^2}sec^2 \theta

Worked Example 2

A ball is thrown from the top of a building of height 20m. The ball is thrown such that the initial speed is U ms-1 and the angle of depression of the throw is 30°. Find:

(a) the horizontal distance travelled in terms of U; and

(b) the horizontal distance when U = 5.

Worked Example 3

A projectile follows the path y=0.3x – 0.1x2. Find the initial speed and the angle of elevation of the particle.

Let’s focus on the direction of motion of a particle at various points along its path. Note that the horizontal component of velocity remains unchanged throughout its motion and that the vertical component of velocity starts with its greatest positive value, reduces to zero by the highest point of its trajectory and then reaches its greatest negative value when it reaches the ground again.

We can calculate the height of a particle at a point on its path if we know the direction it is facing at that point, by relating the horizontal and the vertical components of its velocity at that point and using that relationship with the general equations for the horizontal and vertical components of velocity to calculate the time at which that occurs. Once we know the time, we can plug that into y = utsin \theta - \frac{1}{2}gt^2 to find the height of the particle when it is travelling in that direction.

Worked Example 4

A particle is projected from the top of a tower that is 50m tall. Its initial speed is 25ms-1 at angle of depression 10°. Find the height of the particle above the ground when the downward angle of the direction of the particle is 30°.

Exercise 2

Answers to Exercise 2

  1. 7.52m
  2. 20m
  3. 10.0m
  4. 39.9ms-1
  5. 30.9° below the horizontal
  6. (a) 0.77s, (b) 18.1ms-1
  7. 27.5m
  8. Proof
  9. 45°, 71.6°
  10. Proof

Mixed Exercise

Answers to Mixed Exercise

  1. (i) Proof, (ii) 38.4m, 17.8m, (iii) See image below

2. (i) y = x - \frac{1}{10}x^2 , (ii) x = 4.23, (iii) 𝜽 = 8.8°

3. (i) k = 1/2, 2, proof, (ii) t = 0.894s (iii) 16.3ms-1, 15.9 above horizontal

Worked Exam Style Question

%d bloggers like this: