Question

A 2.0-m-tall basketball player is standing on the floor a distance d from the basket, as shown in the figure below. He shoots the ball at an angle of 40.0o above the horizontal with a speed of 10.67 m/s and makes the shot. The height of the basket is 3.05 m. What is d

please make sure its detailed.

Answer 1

components of the initial velocity is given as :

on horizontal, V_ix = V_i Cos 40⁰                                  { eq. 1 }

where, V_i = 10.67 m/s

V_ix = (10.67 m/s) (0.766)

V_ix = 8.17 m/s

on vertical, V_iy = V_i Sin 40                                  { eq. 2 }

V_iy = (10.67 m/s) (0.642)

V_iy = 6.85 m/s

using equation of motio,

V_ix² = V_iy² + 2 as                            { eq. 3 }

inserting the values in above eq.

(8.17)² = (6.85)² + 2 (9.8) s

66.74 = 46.92 + 19.6 s

19.6 s = 19.82 m

s = 10 m