A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1 m/s, and it travels a distance of 4.6 m. What was (a) the initial direction of the ball and (b) its time of flight?

Answer 1

Use this relationship between distance travelled (D) and angle (A):

D = 2(V^2/g)sinA cos A
= (V^2/g)sin(2A)

You know V, g, and D and so can solve for the angle 2A. Then divide 2A by 2. That will give you the angle A.

For the time of flight T,
V cos A*T = 4.6 m

Answer 2

A) 32 degrees
B)0.76 seconds