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?
+33614 "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
.
