Question

A golf ball hit from the tee at ground level lands 62 m ahead 3.0 s later. What was the initial speed of the ball and at what angle with the horizontal did it come out?

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the golf ball = V

Initial horizontal velocity of the golf ball = V_x

Initial vertical velocity of the golf ball = V_y

Time period the golf ball is in the air = T = 3 sec

Horizontal distance covered by the golf ball = R = 62 m

There is no horizontal force acting on the golf ball therefore the horizontal velocity of the golf ball remains constant.

R = V_xT

62 = V_x(3)

V_x = 20.667 m/s

The golf ball is hit from the ground level and will land at ground level therefore the final vertical displacement of the golf ball is zero therefore,

V_y = 14.715 m/s

V = 25.37 m/s

Angle with the horizontal the golf ball came out at =

= 35.45^o

a) Initial speed of the golf ball = 25.37 m/s

b) Angle with the horizontal the golf ball came out at = 35.45^o