The height, in feet, of a stone thrown vertically upward from 44 feet above ground level with an initial velocity of 40 ft/s is h = -16t^2 + 40t + 44. What is the velocity of the stone when it is 20 feet above the ground?

Answer 1

-16t^2 + 40t + 44 = 20
16 t^2 -40 t -24 = 0
2 t^2 -5 t - 3 = 0
( t-3)(2t+1) = 0
t = 3 seconds
v = dh/dt = -32 t+ 40 = -32(6)+40 = -192 + 40