State the horizontal acceleration and the vertical acceleration of a projectile when air resistance is ignored.
Explain what is meant by saying the horizontal and vertical motions are independent.
A ball is thrown horizontally at 12 m s−1 from a height of 15 m. Find (a) the time to land, (b) the horizontal distance travelled, and (c) the speed and direction with which it hits the ground.
A projectile is launched at 20 m s−1 at 40 degrees above the horizontal over level ground. Find (a) the horizontal and vertical components, (b) the time of flight, (c) the range, and (d) the maximum height.
For the projectile in Section C, find the speed and direction of the projectile 1.0 s after launch.
Describe how the trajectory of the projectile in Section C would change if air resistance were significant. Comment on the range, the maximum height and the symmetry of the path.
Total: 22 marks. Original work by the TheLucidSTEM team. Written in the style of the papers; no past paper question is reproduced.
Answer key · full worked solutionsclick to reveal
horizontal acceleration is zero; vertical acceleration is g (about 9.81 m s−2) downward.
the horizontal and vertical motions do not affect each other; they proceed separately and are linked only by the shared time t.
(a) 15 = ½ × 9.81 × t², so t² = 3.06 and t = 1.75 s (to 2 sf, 1.7 s).
(b) horizontal distance = 12 × 1.75 = 21 m.
(c) vy = 9.81 × 1.75 = 17.2 m s−1; vx = 12 m s−1; resultant = √(12² + 17.2²) = 21 m s−1 at tan−1(17.2 / 12) = 55° below the horizontal.
(a) horizontal = 20 cos 40° = 15.3 m s−1; vertical = 20 sin 40° = 12.9 m s−1.
(b) time to the top = 12.9 / 9.81 = 1.31 s, so time of flight = 2.6 s.
(c) range = 15.3 × 2.62 = 40 m.
(d) maximum height = (12.9)² / (2 × 9.81) = 8.4 m.
horizontal component = 15.3 m s−1 (constant); vertical component = 12.9 − 9.81 × 1.0 = 3.1 m s−1 (still rising).
speed = √(15.3² + 3.1²) = 15.6 m s−1; direction = tan−1(3.1 / 15.3) = 11° above the horizontal.
air resistance opposes the motion, so it reduces both the range and the maximum height. It also makes the path asymmetric: the descent is steeper than the ascent, and the projectile lands at a steeper angle and a lower speed than the symmetric case predicts.