A basketball is thrown at 8.00m/s [up] from a height of 2.00m and eventually comes back down to hit the floor. a)assuming a uniform acceleration of 9.80m/s^2 [down], how long was the ball in flight? b) calculate the balls velocity immediately prior to hitting the floor?

Answer 1

free-fall equation
... h = 1/2 g t^2 + v0 t + h0

0 = -4.9 t^2 + 8 t + 2

a) solve the quadratic for t (you want the positive value)

b) 1/2 m V^2 = 1/2 m v^2 + m g h
... V^2 = v^2 + 2 g h
... V^2 = 8^2 + (2 * 9.8 * 2)

Answer 2

I don't get this. I need to use one of the five key equations for motion with uniform acceleration.