An object is thrown upward from the top of 64-foot ​building with an initial velocity of 48 feet per second. When will the object hit the​ ground?

Question

An object is thrown upward from the top of a 64-foot ​building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h = -16t^2 + 48t + 64. When will the object hit the​ ground?

Answer 1

set h=0 and solve for t for time when it hits the ground

-16^2 + 48t + 64 =0, divide by -16, to get

t^2 -3t -4 =0, factor and set each factor =0

(t-4)(t+1) = 0

t = 4 seconds, ignore the negative solution