Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m above the floor at an angle of 60° above the horizontal, what must the initial speed be (in m/s) if it were to go through the basket?
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Ответ:
The answer to your question is vo = 5.43 m/s
Explanation:
Data
distance = d= 5.8 m
height = 3 m
height 2 = 1.7 m
angle = 60°
vo = ?
g = 9.81 m/s²
Formula
hmax = vo²sinФ/ 2g
Solve for vo²
vo² = 2ghmax / sinФ
Substitution
vo² = 2(9.81)(3 - 1.7) / 0.866
Simplification
vo² = 19.62(1.3) / 0.866
vo² = 25.51 / 0.866
vo² = 29.45
Result
vo = 5.43 m/s
Ответ:
C. A. B.
Explanation:
Credit to Lainey in the comments for the answer!