Question

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 7.3 m/s , and it travels a distance of 3.9 m.

A) What was the initial direction of the ball? B) What was its time of flight?

Answer 1

Initial speed of the ball = V₀ = 7.3 m/s

Horizontal distance travelled by the ball = R = 3.9 m

Initial direction of the ball =

Time of flight = t

The ball is thrown and caught at the same level of height.

Vertical displacement = y = 0 m

Initial vertical speed of the ball = V₀Sin

Initial horizontal speed of the ball = V₀Cos

Acceleration in the vertical direction = -g = -9.81 m/s² (Negative as it is in the opposite direction of initial vertical velocity)

For the vertical direction,

y = (V₀Sin)t + (-g)t²/2

0 = (V₀Sin)t - gt²/2

There is no force acting on the ball in the horizontal direction and therefore the horizontal velocity will remain constant.

R = (V₀Cos)t

Sin2 = 0.718

2 = 45.89^o

= 22.945^o

t = 0.58 sec

a) Initial direction of the ball = 22.945^o above the horizontal

b) Time of flight = 0.58 sec