Question

a person standing at the edge of a cliff throws a rock directly upward. it observed that 2 seconds later, the rock is it maximum height and that 5 seconds after it reached the highest point, it hits the ground at the base of the cliff. what is the initial velocity of the rock

Answer 1

v = Vi - 9.81 t
0 = Vi - 9.81 t at top
Vi = 9.81 t = 19.62 meters/second

Maybe get H as well

falls for five seconds now
h = Hi + Vi T - (1/2) g T^2
0 = Hi + 5 Vi - 4.9(25)
Hi = 122.5 - 96 = 26.5 meters