A 2 m tall basketball player is standing at a horizontal distance 7 m away from a basketball hoop. The time taken by the ball to reach the hoop is 1.50 s. The initial velocity of the ball is 9.33 m/s.
The horizontal component of the initial speed is v cos 60° = 0.5 v
The vertical component of the initial speed is v sin 60° = 0.86 v
The player is 7 m away from the basketball hoop.
So, 7 = 0.5 v * t ----(1)
The net height with respect to the player is 3 m - 2 m = 1 m
So, 1 = - g* t²/2 + 0.86 v * t ----(2)
From (1), we have v * t = 14 ----(3)
Using (3) in (2), we have
1 = - 4.9* t² + 0.86 * 14
1 - 12.04 = -4.9 t²
-4.9 t² = -11.04
t² = 2.25
t = 1.50 s
It takes 1.5 s to reach the hoop.
Substituting t in (1), we have v * t = 14
v = 14/1.5 = 9.33 m/s
The initial velocity of the ball is 9.33 m/s.
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