Question

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.3 m/s at an angle of 15 ∘ below the horizontal. It is released 0.70 m above the floor.

Part A What horizontal distance does the ball cover before bouncing?

Answer 1

Solution:-

Initial speed v = 4.3 m/s
Angle θ = 15 o
Height h = 0.7 m
Horizontal distance does the ball cover before bouncing R =?
Now in the vertical direction :

Initial velocity u = v sin θ
                        = 1.11 m / s
Accleration a = 9.8 m/ s2
Distance S = h
From the relation S = ut + ( 1/ 2) gt 2
                         0.7 = 1.11 t + 4.9 t ^2
4.9 t ^2 +1.11 t - 0.7 = 0

On solving this quadratic equation
t =0.275 s
In horizontal direction :

Time t = 0.281 s
Horizontal velocity U = v cos θ
                                 = 4.153 m/ s
∴ R = U t
        = 1.167 m