A rock dropped from a high platform is moving at 24 ms downward when it strikes the ground. Ignore air resistance. How fast was the rock moving when it had fallen only one-fourth of the distance to the ground?

The main formula to find the speed is depending to

x = 1/2 g t²

v can be found by derivative of x

v = dx / dt = 2x (1/2) gt

so v=gt, if the speed is 24 ms, t = 24/9.8=2.4s

for t = 2.4s, the distance is x=d = 1/2x 9.8*2.4²=29.38m

the one fourth of the distance is d'=1/4d=29.38/4=7.05

let's find t

7.05=1/2*9.8*t² implies t=1.19s

and v = gt, so v=9.8*1.19=11.75ms