Questions on projectiles

Question 1

State the two facts about a projectile's velocity components that make the standard model work.

Accepted Answer

The horizontal component stays constant, because no horizontal force acts on the particle. The vertical component changes at 9.8 m s⁻² downwards, exactly as in free fall. The two motions run independently and share nothing but the clock. B1 B1 for the two components.

Question 2

A ball is projected at 20 m s⁻¹ at 40° above the horizontal. Find the initial horizontal and vertical components of its velocity.

Accepted Answer

Horizontal: 20 cos 40° = 15.3 m s⁻¹. Vertical: 20 sin 40° = 12.9 m s⁻¹ upwards. B1 B1 for the horizontal and vertical components. When the angle is measured from the horizontal, cosine goes with the horizontal component. Store both to more figures than you report.

Question 3

A ball is projected at 22 m s⁻¹ at 35° above the horizontal. Taking g = 9.8 m s⁻², find the horizontal and vertical components of its velocity 1 s after projection, and state whether it is rising or falling at that instant.

Accepted Answer

Horizontal: unchanged at 22 cos 35° = 18.0 m s⁻¹. Vertical, up positive: 22 sin 35° − 9.8 × 1 = 12.62 − 9.8 = 2.82 m s⁻¹. The vertical component is still positive, so the ball is rising. B1 for the horizontal component, M1 for u sin 35° − gt, A1 for 2.82 with the conclusion.

Question 4

A ball is thrown with velocity (10i + 7j) m s⁻¹ and moves freely under gravity. Write down its acceleration as a vector, and its velocity at time t seconds as a vector.

Accepted Answer

Gravity is the only force, so a = −9.8j m s⁻². Then v = u + at gives v = 10i + (7 − 9.8t)j m s⁻¹. B1 for a, B1 for v.

Question 5

A stone is thrown horizontally at 12 m s⁻¹ from a cliff 44.1 m high. Taking g = 9.8 m s⁻², find the time to land and the distance from the cliff base.

Accepted Answer

Vertically the stone starts from rest, so 44.1 = ½ × 9.8 × t², giving t² = 9 and t = 3 s. Horizontally the speed never changes, so the distance is 12 × 3 = 36 m. M1 for the vertical equation, A1 for the time, A1 for the distance.

Question 6

A ball is struck at 24.5 m s⁻¹ at 30° above the horizontal from level ground. Find the time of flight and the range.

Accepted Answer

Components: 24.5 sin 30° = 12.25 m s⁻¹ up and 24.5 cos 30° = 21.2 m s⁻¹ across. Take up as positive. Returning to the same level means s = 0, so 0 = 12.25t − 4.9t² and t = 12.25/4.9 = 2.5 s. Range = 21.2176 × 2.5 = 53.0 m. M1 for resolving, M1 for the vertical equation with s = 0, A1 for the time, A1 for the range. The usual loss is quoting the root t = 0, which is the instant of the strike, rather than the second root.

Question 7

A ball is struck at 24.5 m s⁻¹ at 30° above the horizontal from level ground. Find its maximum height above the ground.

Accepted Answer

At the highest point the vertical component of velocity is zero, so 0 = 12.25² − 2 × 9.8 × s and s = 150.0625/19.6 = 7.66 m. M1 for setting the vertical component to zero, dM1 for using v² = u² + 2as, A1 for 7.66 m. The ball is still travelling at 21.2 m s⁻¹ horizontally up there, so it is the vertical component, not the whole velocity, that is zero there.

Question 8

A ball is thrown from the origin with velocity u = (9i + 14j) m s⁻¹ and moves freely under gravity, so a = −9.8j m s⁻². Find its position vector and its velocity 2 s later, and its speed at that instant.

Accepted Answer

Use the constant acceleration formulae in vector form. s = ut + ½at² = 2(9i + 14j) + ½(−9.8j)(2²) = 18i + 28j − 19.6j = (18i + 8.4j) m: 18 m downrange and 8.4 m up. M1 for s = ut + ½at², A1 for the position. v = u + at = 9i + 14j − 19.6j = (9i − 5.6j) m s⁻¹. M1. The i component is untouched, because the acceleration has none, and the j component has gone negative, so the ball is past its highest point and falling. The speed is √(9² + 5.6²) = √112.36 = 10.6 m s⁻¹. A1 for the velocity and speed.

Question 9

A ball is projected from a point on level ground at 21 m s⁻¹ and lands 40 m away on the ground. Taking g = 9.8 m s⁻², find the two possible angles of projection.

Accepted Answer

Let the angle of projection be A. Components: 21 cos A across, 21 sin A up. Landing means the vertical displacement is zero: 0 = 21t sin A − 4.9t², so the flight time is t = 21 sin A/4.9 = 2 × 21 sin A/9.8. Range = 21 cos A × t = 2 × 21² sin A cos A/9.8 = 21² sin 2A/9.8, using 2 sin A cos A = sin 2A. So sin 2A = 40 × 9.8/441 = 0.889, giving 2A = 62.7° or 117.3°, and A = 31.4° or 58.6°. M1 for the time of flight in terms of A, M1 for the range as the horizontal component times that time, A1 for sin 2A = 0.889, A1 for 31.4°, A1 for 58.6°. Two angles give the same range because sin 2A takes every value between 0 and 1 twice between 0° and 180°: a low trajectory and a high one, symmetric about 45°.

Question 10

A ball is kicked from level ground at 15 m s⁻¹ at 45° above the horizontal, towards a wall 10 m away and 3 m high. Taking g = 9.8 m s⁻², determine whether the ball passes over the wall, and if so by how much.

Accepted Answer

Components: 15 cos 45° = 10.61 m s⁻¹ across and 15 sin 45° = 10.61 m s⁻¹ up. The ball reaches the wall when 10.61t = 10, so t = 0.943 s. Its height then is 10.61 × 0.943 − 4.9 × 0.943² = 10 − 4.36 = 5.64 m, which is above 3 m, so the ball clears the wall by 2.64 m. M1 for the time to reach the wall from the horizontal motion, A1 for t = 0.943, M1 for the height at that time, A1 for 5.64 m with the comparison.

Question 11

A ball is thrown from ground level with velocity (12i + 16j) m s⁻¹ and moves freely under gravity, so a = −9.8j m s⁻². Find the time at which the ball is moving at 45° below the horizontal, and its speed at that instant.

Accepted Answer

v = 12i + (16 − 9.8t)j. Moving at 45° below the horizontal means the vertical component is negative and equal in size to the horizontal one: 16 − 9.8t = −12, so 9.8t = 28 and t = 2.86 s. The velocity is then 12i − 12j, with speed √(12² + 12²) = 12√2 = 17.0 m s⁻¹. M1 for the velocity vector at time t, M1 for setting the j component to −12, A1 for t = 2.86, A1 for 17.0. The sign matters: 16 − 9.8t = 12 gives t = 0.408 s, when the ball is moving at 45° above the horizontal on the way up.

Question 12

A ball is struck at 24.5 m s⁻¹ at 30° above the horizontal from a point 5 m above level ground. Find the time until it lands, the horizontal distance travelled, and its speed on landing.

Accepted Answer

Take up as positive, so the displacement on landing is s = −5. Then −5 = 12.25t − 4.9t², which rearranges to 4.9t² − 12.25t − 5 = 0. The formula gives t = (12.25 + 15.75)/9.8 = 2.86 s. Reject the negative root, which describes a time before the strike. Horizontal distance = 21.2176 × 2.857 = 60.6 m. On landing the vertical velocity is 12.25 − 9.8 × 2.857 = −15.75 m s⁻¹, so the speed is √(21.2176² + 15.75²) = 26.4 m s⁻¹. M1 for s = −5, A1 for the quadratic, M1 A1 for the time, A1 for the horizontal distance, M1 for the vertical velocity on landing, M1 A1 for the resultant speed. Two things are needed. The sign of s has to match the chosen positive direction, and the unrounded 21.2176 has to be carried through, since rounded values carried forward can shift a final answer in the third figure.

Question 13

A sketch shows a projectile's horizontal speed falling away during its flight. Explain why the standard model forbids this, which feature of the acceleration vector says so, and what the only link between the two component motions is.

Accepted Answer

In the model the only force acting is gravity, so the acceleration is a = (0, −g): its horizontal component is exactly zero, and a component of velocity with no acceleration behind it cannot change. The sketch is drawing air resistance, which the model has deliberately left out. B1 for the zero horizontal component of a. The vertical component is the whole of gravity, so the vertical motion is ordinary free fall under suvat, negative because up was chosen as positive. B1 for the vertical component being free fall. The two motions share nothing but the clock: the one quantity in both is t, which is how a vertical statement such as the apex fixes a horizontal one such as the range. B1 for time as the only link.

Question 14

A ball is projected from a point O on level ground with speed 14 m s⁻¹ at 45° above the horizontal. Take O as origin, x horizontal and y vertically upwards, and g = 9.8 m s⁻². By eliminating t between the horizontal and vertical equations, derive the equation of the path in the form y = x − x²/20. Use it to find the horizontal distance at which the ball lands, and the height of the ball when x = 10.

Accepted Answer

Components: 14 cos 45° = 14 sin 45° = 7√2 = 9.90 m s⁻¹. Horizontally x = 7√2 t, so t = x/(7√2). Vertically y = 7√2 t − 4.9t². Substituting, y = x − 4.9x²/(7√2)² = x − 4.9x²/98 = x − x²/20. Landing: y = 0 gives x(1 − x/20) = 0, so x = 20 m. At x = 10, y = 10 − 100/20 = 5 m. M1 for t = x/(u cos 45°), M1 for substituting into the vertical equation, A1 for the given form, A1 for x = 20, M1 for substituting x = 10, A1 for 5 m. Keeping 7√2 exact makes (7√2)² = 98 exact. The equation is a quadratic in x, so the path is a parabola, symmetric about x = 10, which is why the height at x = 10 is also the greatest height of the flight.

Question 15

A particle is projected from level ground with speed u at 45° above the horizontal. Show that its greatest height is u²/(4g) and that its range on the level ground is u²/g. Find the speed of projection needed for a range of 60 m, taking g = 9.8 m s⁻².

Accepted Answer

Both components are u cos 45° = u sin 45° = u/√2. Greatest height, where the vertical velocity is zero: 0 = (u/√2)² − 2gH, so H = u²/(2 × 2g) = u²/(4g). Time of flight, where the vertical displacement is zero: 0 = (u/√2)t − ½gt², so t = 2u/(√2 g) = √2 u/g. Range = horizontal component × time = (u/√2)(√2 u/g) = u²/g. For 60 m, u² = 60 × 9.8 = 588, so u = 24.2 m s⁻¹ (3 s.f.). M1 for the components, M1 for the vertical equation at the top, A1 for u²/(4g), M1 for the flight time and its product with the horizontal component, A1 for u²/g, A1 for 24.2. At 45° the range is four times the greatest height, and the range grows with the square of the speed, so doubling the speed quadruples the range.