A diver runs horizontally with a speed of 1.30 m/s off a platform that is 8.6 m above the water. What is his speed just before striking the water?

Answer 1

8.6 = 4.9 t^2
t = 1.32 seconds to fall 8.6 meters
v = g t = 9.81*1.32 = 13 m/s
so
speed = sqrt (169 + 1.3*2) =sqrt(169+1.69)
=13.06 m/s