A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the velocity just before it hits the water?

Answer 1

Hi = bridge height
0 = water height
a = g = -9.81 m/s^2
v = Vi - g t = 11 -9.81 t
0 =Hi + Vi t + (1/2) a t^2 = Hi + 11 t - 4.9 t^2
but t = 5.5
v = 11 - 9.81(5.5) = 11 - 54 = - 43 m/s

Answer 2

I assume there is a part b asking for Hi, the bridge height.