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
