Projectile – SS2 Further Mathematics Lesson Note (Second Term)

Projectile motion is the motion of an object that follows a curved, parabolic path. It results from two independent motions at right angles to each other: a constant horizontal velocity and a vertical free fall under gravity. A stone released from a catapult and a bullet fired from a gun are examples.

Behavioural objectives

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

1. explain what a projectile is and name the two motions that make it up;
2. resolve the initial velocity into horizontal and vertical components;
3. derive and use the formulas for time of flight, maximum height and range;
4. state the angle that gives the maximum range;
5. solve simple problems on projectiles.

What is projectile motion?

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

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

Examples of projectile motion are the motion of:

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

If a particle is projected with velocity U at an angle θ to the horizontal:

U_y = U sin θ (vertical component) … (1)

U_x = U cos θ (horizontal component) … (2)

What terms are associated with projectile motion?

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, where v = 0 and a = -g for the vertical motion:

0 = U sin θ – gt

t = U sin θ / g … (3)

T = 2t = 2U sin θ / g … (4)

Maximum height (H)

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

Using v² = U² sin²θ – 2gH, and v = 0 at maximum height:

2gH = U² sin²θ

H = U² sin²θ / 2g … (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, the acceleration is zero, so S = ut + ½at² becomes S = ut. Therefore:

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. So:

R = U² sin 2θ / g

For maximum range, θ = 45°, so sin 2θ = sin 90° = 1. Therefore:

R_max = U² / g

What are the uses of projectiles?

1. To launch missiles in modern warfare.
2. To give athletes maximum take-off 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: the 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.

Solution

Horizontal velocity of bomber = 60 km/min = 10³ m/s = 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, i.e. 1000 m/s.

Considering the vertical motion of the bomb, with u = 0:

h = ut + ½gt² = ½gt²

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

3000 = ½ × 10 × t²

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 = 1000 × 10√6 = 2.449 × 10⁴ m

tan θ = s / 3000 = 2.449 × 10⁴ / 3000

θ = 83.02°

Example 2: the catapult

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

1. the time of flight
2. the maximum height attained
3. the range

Solution (take g = 10 m/s²)

(a) The time of flight

T = 2U sin θ / g = (2 × 30 × sin 60°) / 10

T = 5.2 s

(b) The maximum height

H = U² sin²θ / 2g = (30² × sin² 60°) / 20

H = 33.75 m

(c) The range

R = U² sin 2θ / g = (30² × sin 120°) / 10

R = 90 sin 120°

R = 77.9 m

Evaluation

1. What is a projectile? Name the two independent motions that make it up.
2. Write the formulas for the time of flight, the maximum height and the range of a projectile.
3. Show that the range is greatest when θ = 45°, and state the maximum range.
4. A ball is thrown with an initial velocity of 20 m/s at an elevation of 30°. Find the time of flight. (Take g = 10 m/s².)

Assignment

A stone is projected with an initial velocity of 40 m/s at an elevation of 45°. Taking g = 10 m/s², find:

1. the time of flight
2. the maximum height attained
3. the range