On top of a hill, a rocket is launched from a distance 80 feet above a lake.

Question

On top of a hill, a rocket is launched from a distance 80 feet above a lake. The rocket will fall into the lake after its engine burns out. The rocket's height, h, in feet above the surface of the lake, is given by the equation, h = -16t 2 + 64t + 80, where t is time in seconds

 

The maximum height of the rocket is _ feet.

Answer 1

h(t) = -16t^2 + 64t + 80

 

This is a problem of vertical motion under constant acceleration.  The equation is of the form:

 

h(t) = (1/2)at^2 + v0*t + h0, a = constant acceleration, v0 = initial vertical speed, h0 = initial height

 

By inspection, a = -32 ft/s^2, v0 = 64 ft/s, h0 = 80 ft

 

Maximum height is attained when v(t) is instantaneously zero.  Under constant acceleration,

 

v(t) = at + v0

 

Set v(t) = 0, solve for t, then plug t into h(t) to get maximum height.

 

at + v0 = 0

 

t = -v0 / a = -(64) / (-32) s = 2 s

 

h(2) = [-16(2^2) + 64(2) + 80] ft = 144 ft

 

h_max = 144 ft