An object is launched from a platform The height of its moti

An object is launched from a platform. The height of its motion in feet, h, for each second, t, after the launch, is given by h = -t^2 + 4t + 12 a. How many seconds does it take for the object to hit the ground? b. What is the maximum height that the object reaches?

Solution

When object reaches ground h=0

-t^2+4t+12=0

t^2-6t+2t-12=0

(t-6)(t+2)=0

t=6sec

Max height is differentiation of f(t)

2t-4=0

t=2

Max ht= -4+8+12=16