Projectile and its Implications – SS2 Physics Lesson Note (First Term)

A projectile moves along a curved, parabolic path. The motion is made up of two independent motions at right angles to each other: a constant horizontal velocity and a vertical free fall due to gravity. The time of flight, maximum height and range of a projectile depend on its initial velocity U and angle of projection θ.

Behavioural objectives

By the end of the lesson, students should be able to:

• describe projectile motion and give examples
• resolve the initial velocity into horizontal and vertical components
• derive and use the formulae for time of flight, maximum height and range
• state the angle of projection that gives the maximum range
• state some uses of projectiles
• solve numerical problems on projectiles

What is projectile motion?

A projectile motion follows a curved or parabolic path. It is due to two independent motions at right angles to each other. These motions are:

1. horizontal constant velocity
2. vertical free fall due to gravity

Examples of projectile motion

1. A thrown rubber ball bouncing from a wall
2. An athlete doing the high jump
3. A stone released from a catapult
4. A bullet fired from a gun
5. A cricket ball thrown against a vertical wall

Components of the initial velocity

Uy = U sin θ (vertical component) (equation 1)

Ux = U cos θ (horizontal component) (equation 2)

What terms are associated with projectiles?

Time of flight (T)

The time of flight of a projectile is the time required for it to return to the same level from which it is projected.

Let t be the time to reach the greatest height. Using v = u + at, with v = 0 and a = −g:

0 = U sin θ − gt

t = U sin θ / g (equation 3)

T = 2t = 2U sin θ / g (equation 4)

Maximum height (H)

The maximum height is the highest vertical distance measured from the horizontal projection plane.

For maximum height H:

V² = U² sin²θ − 2gH

At maximum height, V = 0, so 2gH = U² sin²θ

H = U² sin²θ / 2g (equation 5)

Range (R)

The range is the horizontal distance from the point of projection of a particle to the point where the particle hits the projection plane again.

Horizontally, using S = ut + ½at² (where a = 0 for the horizontal motion), or:

S = R = U cos θ × t (distance = velocity × time, where the time is the time of flight)

R = U cos θ × (2U sin θ / g)

R = 2U² sin θ cos θ / g

From trigonometry, 2 sin θ cos θ = sin 2θ. Therefore:

R = U² sin 2θ / g

For maximum range, θ = 45°, so sin 2θ = sin 2(45°) = sin 90° = 1.

Rmax = U² / g

What are the uses of projectiles?

1. To launch missiles in modern warfare
2. To give athletes maximum takeoff speed at meets
3. In artillery warfare, to strike a specified target, the bomb must be released when the target appears at the angle of depression φ given by: tan φ = (1/u)√(gh/2)

Worked examples

Example 1: bomb released from a bomber

A bomber on a military mission is flying horizontally at a height of 3000 m above the ground at 60 km min⁻¹. It drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released.

Horizontal velocity of the bomber = 60 km/min = 10³ m s⁻¹ = 1000 m/s.

The bomb falls with a vertical acceleration g = 10 m/s². At the release of the bomb, it moves with a horizontal velocity equal to that of the aircraft, that is 1000 m/s.

Considering the vertical motion of the bomb, h = ut + ½gt² (u = 0), so h = ½gt².

Here t is the time the bomb takes to reach the ground:

3000 = ½gt²

t² = 600

t = 10√6 s

Considering the horizontal motion, the horizontal distance moved by the bomb in time t is:

s = horizontal velocity × time

s = 1000 × 10√6

s = 2.449 × 10⁴ m

But tan θ = s / 3000 = 2.449 × 10⁴ / 3000

θ = 83.02°

Example 2: stone shot from a catapult

A stone is shot out from a catapult with an initial velocity of 30 m/s at an elevation of 60°. Find:

• the time of flight
• the maximum height attained
• the range

i. The time of flight

T = 2U sin θ / g

T = (2 × 30 × sin 60°) / 10

T = 5.2 s

ii. The maximum height

H = U² sin²θ / 2g

H = 30² sin²(60°) / 20

H = 33.75 m

iii. The range

R = U² sin 2θ / g

R = 30² sin 2(60°) / 10

R = 90 sin 120°

R = 77.9 m

Evaluation

1. Name the two independent motions that make up projectile motion.
2. Define time of flight, maximum height and range.
3. Show that the range of a projectile is R = U² sin 2θ / g.
4. At what angle of projection is the range greatest, and what is the maximum range in terms of U and g?

Assignment

A stone is shot from a catapult with an initial velocity of 20 m/s at an elevation of 30°. Taking g = 10 m/s², find the time of flight, the maximum height attained and the range.