Question

A basketball is shot by a player at a height of 2.0m. The initial angle was 53° above the horizontal. At the highest point, the ball was travelling 6 m/s. If he scored (the ball went through the rim that is 3.00m above the ground), what was the player's horizontal distance from the basket?

Answer 1

At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that

v cos(53°) = 6 m/s ==> v = (6 m/s) sec(53°) ≈ 9.97 m/s

The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are

x = (9.97 m/s) cos(53°) t = (6 m/s) t

y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

Find the times t for which the ball reaches a height of 3.00 m:

3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²

==> t ≈ 0.137 s or t ≈ 1.49 s

The second time is the one we care about, because it's the one for which the ball would be falling into the basket.

Now find the distance x traveled by the ball after this time:

x = (6 m/s) (1.49 s) ≈ 8.93 m