Question

A bottle rocket is launched upward at 128 feet per second from a platform that is 136 feet high. Its path can be mapped by the equation h = -16t^2 + v'o't + h'o (where h=height, t=time, v'o=initial velocity, and h'o=initial height).

Which of the following equations represents the height of the rocket in vertex form?

A. h=-16(t-8)^2+136
B. h=-16(t^2-8t)+136
C. h=16(t-4)^2
D. h=-16(t-4)^2+392

Answer 1

d2x/dt2=-g (which we'll approximate as a constant -32ft/s^2)

d2x/dt2=-32 integrating we get:

dx/dt=-32t+C, where C is the initial velocity which we are told is 128ft/s so

dx/dt=-32t+128, integrating again we get:

h(t)=-16t^2+128t+C, where C is the initial height of the object..which we are told is 136 ft so

h(t)=-16t^2+128t+136 if we factor the first two terms we get:

h=-16(t^2-8t)+136