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

#1**+1 **

"*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: v^{2} = u^{2} + 2as where:

v = final vertical velocity (0 m/s)

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

a = acceleration ( -9.8 m/s^{2})

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