Questions on projectile motion

Question 1

A projectile is launched horizontally. Explain why its horizontal and vertical motions can be treated separately.

Accepted Answer

Gravity acts only vertically, so it changes the vertical velocity but not the horizontal velocity (1). The horizontal motion is at constant velocity and the vertical motion is free fall; the two are independent and share only the time (1).

Question 2

A ball rolls off a table 1.2 m high with a horizontal velocity of 3.0 m s⁻¹. Taking g = 9.81 m s⁻², calculate the time to reach the floor and the horizontal distance travelled.

Accepted Answer

Fall time: t = √(2h/g) = √(2 × 1.2/9.81) (1) t = 0.495 s (1) Horizontal distance: x = 3.0 × 0.495 = 1.48 m (1)

Question 3

State what happens to the horizontal and vertical components of a projectile's velocity during its flight (ignore air resistance).

Accepted Answer

The horizontal component stays constant (1); the vertical component changes uniformly at g, decreasing on the way up and increasing on the way down (1).

Question 4

Ignoring air resistance, state, for a projectile at the highest point of its path, (a) the magnitude of the vertical component of its velocity, and (b) the magnitude and direction of its acceleration.

Accepted Answer

(a) Zero (1). (b) g = 9.81 m s⁻², directed vertically downwards; gravity acts throughout the flight (1).

Question 5

Two identical coins rest at the edge of a bench. At the same instant, one is struck so it leaves the bench horizontally at high speed while the other is nudged off the edge and falls from rest. State and explain which coin lands first. Ignore air resistance.

Accepted Answer

They land at the same time (1). The vertical motions are identical, both falling from rest at g; the horizontal velocity has no effect on the vertical motion (1).

Question 6

A stone is thrown horizontally at 20 m s⁻¹ from the top of a cliff 45 m high. Taking g = 9.81 m s⁻², calculate (a) the time of flight, (b) the horizontal distance travelled, and (c) the vertical component of velocity just before impact.

Accepted Answer

(a) t = √(2 × 45/9.81) (1) t = 3.03 s (1) (b) x = 20 × 3.03 = 60.6 m (1) (c) v_y = gt = 9.81 × 3.03 = 29.7 m s⁻¹ (1)

Question 7

A ball is launched at 25 m s⁻¹ at 30° above the horizontal over level ground. Taking g = 9.81 m s⁻², calculate (a) the horizontal and vertical components of the initial velocity, (b) the time of flight, and (c) the horizontal range.

Accepted Answer

(a) u_x = 25 cos 30° = 21.7 m s⁻¹ (1) u_y = 25 sin 30° = 12.5 m s⁻¹ (1) (b) T = 2u_y/g (1) T = 2.55 s (1) (c) R = u_xT = 55.2 m (1)

Question 8

For the launch in the previous question, calculate the maximum height reached above the launch point.

Accepted Answer

H = u_y²/2g (1) H = 12.5²/(2 × 9.81) (1) H = 7.96 m (1)

Question 9

In a film stunt, a cyclist rides horizontally off a flat roof 3.2 m above the ground and lands 7.2 m from the base of the building. Taking g = 9.81 m s⁻², determine the cyclist's speed on leaving the roof.

Accepted Answer

Fall time from the vertical motion: t = √(2 × 3.2/9.81) = 0.81 s (1). Horizontally the speed is constant: u = x/t = 7.2/0.808 (1) = 8.9 m s⁻¹ (1).

Question 10

A stunt diver runs off the top of a cliff with a horizontal velocity of 12 m s⁻¹ and enters the sea 30 m below. Taking g = 9.81 m s⁻², calculate the magnitude of the diver's velocity on entering the water and the angle it makes with the horizontal.

Accepted Answer

Vertical component at entry: v_y = √(2gh) = √(2 × 9.81 × 30) = 24.3 m s⁻¹ (1). The horizontal component is unchanged at 12 m s⁻¹, so v = √(12² + 24.3²) (1) = 27 m s⁻¹ (1). tan θ = 24.3/12, so θ = 64° below the horizontal (1).

Question 11

A ball is thrown at 24 m s⁻¹ at 55° above the horizontal. Taking g = 9.81 m s⁻², calculate the horizontal and vertical components of its velocity 1.5 s after launch, and state whether the ball is still rising.

Accepted Answer

Horizontal: unchanged at 24 cos 55° = 13.8 m s⁻¹ (1). Vertical: v_y = 24 sin 55° − 9.81 × 1.5 = 4.9 m s⁻¹ (1). v_y is still positive (upwards), so the ball is still rising (1).

Question 12

A projectile is launched over level ground at 30 m s⁻¹ at 40° to the horizontal. Using R = u² sin 2θ / g with g = 9.81 m s⁻², calculate its range and state the launch angle that would give the maximum range.

Accepted Answer

2θ = 80° (1) R = 30² × sin 80° / 9.81 (1) R = 90.3 m (1) The maximum range for a given speed occurs at 45° (1)

Question 13

A ball is launched at 15 m s⁻¹ at 25° above the horizontal from a point 2.0 m above the ground. Taking g = 9.81 m s⁻², determine the time of flight until it lands and the horizontal distance travelled.

Accepted Answer

u_x = 13.6 m s⁻¹ (1) u_y = 6.34 m s⁻¹ (1) Vertical: 0 = 2.0 + u_yt − ½gt² (1) t = 1.55 s (1) Range = u_xt = 21.1 m (1)

Question 14

Describe how air resistance affects the trajectory, the maximum height and the range of a projectile, compared with the motion when air resistance is ignored.

Accepted Answer

Air resistance opposes the motion, reducing both horizontal and vertical speeds (1). The path becomes asymmetric, steeper on the way down (1). The maximum height and range are reduced, and the angle for maximum range falls below 45° (1).

Question 15

A javelin leaves the thrower's hand at 26 m s⁻¹ at 40° above the horizontal. Calculate the javelin's speed at the highest point of its flight, and explain why the speed there is not zero.

Accepted Answer

At the highest point the vertical component of velocity is zero, but the horizontal component is unchanged throughout the flight (1). Speed = 26 cos 40° (1) = 19.9 m s⁻¹ (1).

Question 16

A rugby conversion is kicked from a point 43 m from the posts. The ball leaves the ground at 22 m s⁻¹ at 38° above the horizontal, travelling straight towards the posts. To score, the ball must pass above the crossbar, which is 3.0 m above the ground. Ignoring air resistance and taking g = 9.81 m s⁻², deduce whether the kick scores.

Accepted Answer

Components: u_x = 22 cos 38° = 17.3 m s⁻¹, u_y = 22 sin 38° = 13.5 m s⁻¹ (1). Time to reach the posts: t = 43/17.34 = 2.480 s (1). Height then: y = u_yt − ½gt² = 13.54 × 2.480 − 4.905 × 2.480² (1) = 3.4 m (1). 3.4 m is above the 3.0 m crossbar, so the kick scores, clearing it by about 0.4 m (1).

Question 17

An aircraft on a flood-relief flight travels horizontally at 65 m s⁻¹ at a height of 300 m, aiming to drop a supply package onto a marked target. Ignoring air resistance and taking g = 9.81 m s⁻², calculate the horizontal distance before the target at which the package must be released, and state where the aircraft is relative to the package at the moment it lands.

Accepted Answer

Fall time: t = √(2 × 300/9.81) = 7.82 s (1). Release distance = 65 × 7.82 (1) = 508 m ≈ 510 m before the target (1). The aircraft is directly above the package when it lands: both keep the same constant horizontal velocity of 65 m s⁻¹ (1).