Question

A rocket is fired straight up from a 60 ft. platform with an initial velocity of 96 ft/sec. The height of the rocket, h(t), is found using the function h\left( t \right) = - 16{t^2} + 96t + 60 where t is the time in seconds.
Find the maximum height.

Answer 1

h(t) = -16t^2 + 96t + 60
v(t) = -32t + 96
at max height , v = 0
-32t + 96=0
32t = 96
t = 96/32 = 3

h(3) = -16(9) + 96(3) + 60 = 204

or

h(t) = -16(t^2 - 6t + 9-9) + 60
= -16( (t-3)^2 - 9) + 60
= -16(t-3)^2 + 144 + 60
= -16(t-3)^2 + 204

this has a vertex of (3,204)
thus a max of 204 when t - 3

Answer 2

3