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?
"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
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