Projectile - Question 2

A golf ball is struck from level ground at 25 ms^-1 at 40° above the horizontal. Ignoring air resistance, calculate (i) the maximum height, (ii) the time of flight and (iii) the horizontal range.Step 1: resolve the launch velocity.u_H = 25cos40° = 19.15 ms^-1    u_V = 25sin40° = 16.07 ms^-1The horizontal and vertical motions are independent. Horizontally there is no force, so u_H stays constant. Vertically the acceleration is 9.81 ms^-2 downwards, so take upwards as positive and use a = -9.81 ms^-2.(i) Maximum height. At the top the vertical velocity is zero. Using v² = u² + 2as:0 = 16.07² + 2(-9.81)s   so   s = 13.2 m(ii) Time of flight. Using v = u + at, the time to reach the top is t = 16.07 ÷ 9.81 = 1.638 s. On level ground the path is symmetrical, so the time of flight is 2 × 1.638 = 3.28 s.(iii) Range. Horizontally the velocity is constant, so s = ut:s = 19.15 × 3.276 = 62.7 mKeep unrounded values in your calculator until the final answer. Rounding u_H to 19.2 ms^-1 and t to 3.28 s would give 63.0 m, which could cost the accuracy mark.