A ball is thrown horizontally from a height of 24.93 m and hits the ground with a speed that is 3.0 times its initial speed. What was the initial speed?

Answer 1

Y^2 = Yo^2 + 2g*h
Y^2 = 0 + 19.6*24.93 = 488.6
Y = 22.1 m/s.

Xo = 22.1/3 = 7.37 m/s.=Initial velocity