A clump of soft clay is thrown horizontally from 78.40 m above the ground with a speed of 27.0 m/s. Where is the clay after 4.5 s? Assume it sticks in place when it hits the ground.

Question

horizontal ....m from the launch position
vertical ....m from the ground

Answer 1

With no initial vertical velocity component, it hits the ground after time T given by
78.4 = (g/2) T^2
T = 4.00 s

It's horizontal location is the initial (horizontal) velocity muliplied by that time. It does not move after that.