A free throw line in basketball is 4.57 m (15ft.) from the basket, which is 3.05

Question

A free throw line in basketball is 4.57 m (15ft.) from the basket, which is 3.05 m (10ft) above the floor. A player standing on the free throw line throws the ball with the initial speed of 8.15 m/s, releasing it at height of 2.44 m (8ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved on solving projectile motion problems. A free throw line in basketball is 4.57 m (15ft.) from the basket, which is 3.05 m (10ft) above the floor. A player standing on the free throw line throws the ball with the initial speed of 8.15 m/s, releasing it at height of 2.44 m (8ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved on solving projectile motion problems.

Explanation / Answer

Use the trajectory equation:
y = h + x·tan - g·x² / (2v²·cos²)
y = 3.05 m
h = 2.44 m
x = 4.57 m
= ?
v = 8.15 m/s
3.05 = 2.44 + 4.57tan - 9.8(4.57)² / (2(8.15)²cos²)
0 = -0.61 + 4.57tan - 1.54sec² since sec = 1/cos
But sec² = 1 + tan², so
0 = -0.61 + 4.57tan - 1.54 - 1.54tan² = -2.14 + 4.57tan - 1.54tan²
quadratic in tan; roots at tan = 0.58, 2.38
taking "the large initial angle",
= arctan2.38 = 67º

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