A golfer hits the ball, imparting an initial velocity, u, of 40 m.s-1 to the ball, which departs at an angle α of 35º from the horizontal. Disregard any possible effects of air resistance, and assume that the ball will land at the same height as where it took off. (g = 9.80 m.s-2)

How high will the ball travel?

The greatest height of the trajectory, H = u² sin²α₀ / 2g = 40² (m².s²) x 0.574²/2 x 9.80(m.s^-2) = 26.9 m.

How long will the ball take to reach the top of its trajectory?

The time taken is T = u sinα ₀ / g = 40 (m.s^-1 x 0.574 / 9.80 (m.s^-2) = 2.34 s.

How far will the ball travel?

The range, R = u² sin2α₀ / g = 40² (m².s²) x sin(2 x 35º) / 9.80 (m.s^-2) = 153 m.

How long will the ball move in the air?

The time of flight will be twice the time taken to reach the maximum height of the trajectory = 4.68 s.

What is the speed of the ball on landing?

In the absence of drag, the speed of the ball on landing will be the same as its initial speed, 40 m.s^-1. The velocity will not be the same, as the ball will be travelling in a different direction.