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A basketball is shot at 65 degrees above horizontal with an initial velocity of 8.8m/s. The basketball hoop is 2 meters above where the ball left the persons hands. What is the maximum height the ball reached above its starting point?

Guest Aug 15, 2017
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"A basketball is shot at 65 degrees above horizontal with an initial velocity of 8.8m/s. ... What is the maximum height the ball reached above its starting point?"

 

Using the constant acceleration equation:  v2 = u2 + 2as where: 

 

v = final vertical velocity  (0 m/s)

u = initial vertical velocity (8.8*sin(65°) m/s)

a = acceleration ( -9.8 m/s2)

s = vertical distance travelled

 

0 = [8.8*sin(65°)]2 - 2*9.8*s

 

s = [8.8*sin(65°)]2 /(2*9.8) or s = 3.245 m

.

Alan  Aug 15, 2017

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