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.

Answer 1

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