A toy rocket is launched straight up

Question

A toy rocket is launched straight up from the roof of a garage with an initial velocity of 104 feet per second. The height  of the rocket in feet, at  seconds after it was launched, is described by h(t)=-16t^2+104t+18 .

Find the maximum height of the rocket. 

Answer 1

h(t) = -16t^2 + 104t + 18 = height at time t

max height is when velocity = 0 = h'(t) = -32t +104 = 0

t = 104/32 = 52/16 = 26/8 = 13/4 = 3.25 seconds to reach max height

plug that into h(t)

h(13/4) = - 16(13/4)^2 + 104(13/4) + 18 = -169+26(13) +18 = 26(13) - 151 = 338-151 = 187 feet = max height

that's the calculus method

or try algebra, put the height equation in vertex form, complete the square

-16t^2 +104t + 18

= -16(t^2 -26t+169/16)+18+169

=-16(t-13)^2 + 187 vertex is (13, 187) y coordinate of vertex is max height = 187 feet

Answer 2

v(t) = h'(t) = - 32t + 104 = 0; t = 104/32 = 3.25 s

max h = h(3.25) = - 16·3.25² + 104·3.25 + 18 = 187 ft

Answer 3

The graph is a downward parabola with a peak at its apex. The maximum height is the y value at the apex.

The apex for the equation ax² + bx + c = y occurs at x = -b/2a

You can find the time to the peak (apex), then plug that into the h(t) equation to find h_max.

Please consider a tutor. Take care.