Question

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.2 m/s at an angle of 15 ∘above the horizontal. It is released 0.85 m above the floor.

What horizontal distance does the ball cover before bouncing?

Express your answer using two significant figures.

D = ? m

Answer 1

In horizontal direction:

v = 4.2 * cos 15

= 4.2*0.966 m/s

= 4.06 m/s

Now use:

D = v*t since there is no acceleration is horizontal direction

to calculate t, consider vertical motion:

vi = 4.2*sin 15

= 4.2 * 0.259 m/s

= 1.09 m/s

d = -0.85 m

a = -9.8 m/s^2

use:

d = vi*t + 0.5*a*t^2

-0.85 = 1.09*t + 0.5*(-9.8)*t^2

4.9*t^2 - 1.09*t - 0.85 = 0

This is quadratic equation (at^2+bt+c=0)

a = 4.9

b = -1.09

c = -0.85

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 17.85

roots are :

t = 0.5423 and t = -0.3199

since t can't be negative, the possible value of t is

t = 0.5423

Use:

D = v*t

= 4.2 * 0.5423

= 2.28 m

Answer: 2.3 m