A basketball player 1.95 m tall wants to make a basket from a distance of 12 m. The hoop is at a height of 3.05 m. If he shoots the ball (from a height of 1.95 m) at a 61.5◦ angle, at what initial speed must he shoot the basketball so that it goes through the hoop without striking the back- board? The acceleration due to gravity is 9.8 m/s2 . Neglect air friction.
Answer in units of m/s
Answer in units of m/s
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y = -g*x²/(2V²cos²Θ) + x*tanΘ + yi
Rearranging,
V = (x/cosΘ)*√[-g/(2(y - xtanΘ - yi))]
In your problem,
x - 12, Θ = 61.5°, y = 3.05, yi = 1.95
V = 12.15 m/s
Check my substitution math!
Rearranging,
V = (x/cosΘ)*√[-g/(2(y - xtanΘ - yi))]
In your problem,
x - 12, Θ = 61.5°, y = 3.05, yi = 1.95
V = 12.15 m/s
Check my substitution math!