Projectile Motion

Projectile Motion - National 5 Recap

When an object is thrown/launched/fired on the Earth's surface, it will experience a downwards Force due to Gravity. The path the object takes through the air is a combination of the forward motion and the downwards motion. This results in the projectile (object being projected) following a curved path through the air.

Each of these motions are independent of each other, leading to some counterintuitive results. For example, If a bullet cartridge drops out of a gun at the instant the gun is fired, both the bullet and the cartridge will hit the ground at the same time.

The video below shows the "Monkey and the Hunter" thought experiment based upon Projectile Motion theory :-

For all Projectile Motion :-

1. Constant Horizontal Velocity - No unbalanced Forces acting

2. Constant Vertical Acceleration - Acceleration due to Gravity ( a = 9.8 ms-2 )


Note - At Higher level it is assumed there is no air resistance, allowing constant horizontal Velocity.

In the Higher Physics course, the understanding above can be combined with the equations of motion to calculate many different aspects of Projectile motion.


Horizontal Projectiles

If an object is projected in a horizontal direction, the object will follow a curved path as shown above. The starting conditions are as follows :-


1. Zero Vertical Velocity

2. Non-Zero Horizontal Velocity


By identifying all known variables in each direction, the equations of motion can then be applied to find unknown variables.

Example 1 -

A ball is kicked horizontally from the top of a cliff as shown below:-

If its initial Horizontal velocity is 10 ms-1, and it takes 4 seconds to hit the ground, calculate the following :-

1. How far from the base of the cliff does the ball land.

2. The vertical Velocity at the moment of Impact.


The Height of the cliff

Step 1 - Identify known information in each plane

Horizontally :-

vh = 10 ms-1

t = 4 s

sh = ?

Vertically :-

uv = 0 ms-1

vv = ?

a = - 9.8 ms-2

sv = ?

t = 4 s


Step 2 - Calculate in each plane separately :-

vh = sh / t

sh = vh x t

sh =10 x 4

sh = 40 m

vv = uv + at

vv = 0 + (-9.8 x 4)

vv = -39.2 ms-1

vv = 39.2 ms-1 (downwards)


s = ut + 0.5 x a t2

s = 0 + (0.5 x -9.8 x 42)

s = -78.4 m

s = 78.4 m (downwards)

Oblique Projectiles

Not all projectiles start off in the horizontal direction. In most cases, a projectile will have a combination of horizontal and vertical Velocities at the start of its motion:-


Note - The above diagram shows the vertical component of the Velocity (red arrows) and the horizontal component of the Velocity (blue arrows) across the motion.

Note - Without air resistance, all oblique projectiles follow a symmetrical path, this means that the time taken to reach the highest point on the path is half of the time of the full motion.

In order to calculate the resultant motion, the initial horizontal and vertical components of the Velocity must be found :-

Once the component Velocities are found, the question can be treated using the exact same method as the horizontal projectile.

The following diagram shows the values of the velocity components at key points in the motion:-


At the maximum height :-

vv = 0

vh = uh


Horizontal Displacement = 1/2 R

Time = 1/2 T

Note - At end of motion the projectile impacts with an identical magnitude velocity that it began with, only now Vv acts in the negative direction. This is why when people fire guns in 'celebration' into the air, people several km away can be killed when the bullets impact at the end of their flight.

Example 2 -

A golf ball is struck with a golf club and leaves the tee with a Velocity of 24 ms-1 at an angle of 60° to the horizontal. Calculate :-

1. The maximum Height of the golf ball.

2. The range of the golf ball.

Identify known information in each plane

Horizontally :-

vh = 24 cos60 = 12 ms-1

t = ?

s = ?


Vertically :-

uv = 24 sin60 = 20.8 ms-1

vv = 0 (at maximum Height)

a = - 9.8 ms-2

s = ?

t = ?


Maximum Height is a vertical quantity, use equations of motion to calculate S vertically.


vv2 = uv2 + 2as

0 = 20.82 + ( 2 x -9.8 x s )

s = 22.1 m ( upwards )


Time is constant for both horizontal and vertical motion, use equations of motion to calculate t vertically to reach maximum Height, then double this value to find time for whole motion. This can then be used to calculate the Range horizontally.

vv = uv + at

0 = 20.8 + ( -9.8 x t )

t = 2.1 s

Total time of flight = 2 x 2.1 = 4.2 s

vh = sh / t

12 = sh / 4.2

Range = sh = 50.4 m

Oblique Projectile - Uneven Heights

The above method works well when the start and end points of the motion are at equal heights to each other. If, however, there is a difference in Height between these points, additional calculation is required.

Example 3 -

A golfer on an elevated tee hits a golf ball with an initial velocity of 35 ms-1 at an angle of 40° to the horizontal. The ball travels through the air and lands at point R. Point R is 12 m below the Height of the tee, as shown.

Calculate :-

1. The Time taken to reach maximum Height (point P).

2. The total horizontal Distance (d) of the ball if it takes 0.48 s to travel Q to R.


Identify known information in each plane

Horizontally :-

vh = 35 cos40 = 26.8 ms-1

t = ?

s = ?


Vertically :-

uv = 35 sin40 = 22.5 ms-1

vv = 0 ( at Maximum Height )

a = - 9.8 ms-2

s = ?

t = ?


Use equations of motion to calculate t vertically to reach maximum Height.

vv = uv + at

0 = 22.5 + (-9.8 x t)

t = 2.3 s (to reach P)


To calculate the total horizontal distance, the time from start to P, P to Q and Q to R must be found before the horizontal distance can be calculated.

The motion is symmetrical from start to Q, therefore if start to P takes 2.3 s, then P to Q takes 2.3 s also. Therefore total Time from start to Q = 2.3 + 2.3 + 0.48 = 5.08 s

Use equations of motion to calculate total horizontal distance.

vh = sh / t

26.8 = sh / 5.08

sh = 136 m