Posted by Perry on .
a ball is thrown upward with an initial velocity of 38.9 m/s from the edge of a cliff of height H. The ball lands on the ground at the base of the cliff 12.8 s after it is thrown.

physics. 
Damon,
I assume you want H.
h = H + Vi t + (1/2)(9.8) t^2 = 0 at ground
0 = H + 38.9 (12.8)  4.9 (12.8)^2
solve for H