Question

A rocket is launched from atop a 76-foot cliff with an initial velocity of 107 ft/s. The height of the rocket
above the ground at time t is given by .
A) When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.
(Solve Algebraically, Hint: Use the Quadratic Formula to solve and the round the square root answer one
decimal place)

B) When will the rocket reach 167ft from the ground?

Answer 1

The height equation that should have been given to you would be ...
height = -16t^2 + 107t + 76
"When will the rocket hit the ground". I will assume by ground you mean 76 below the cliff, so height = 0
-16t^2 + 107t + 76 = 0
16t^2 - 107t - 76 = 0
They told you to use the quadratic equation formula, so ...
t = (107 ± √(107^2 - 4(16)(-76)) )/32
= ..... you do the button pushing, use the positive result only

b)
-16t^2 + 107t + 76 = 167
16t^2 - 107t + 91 = 0
You will get 2 answers, both positive and both exact.
One will be the time it will reach a height of 167 on its way up, and the other
on its way down.

Answer 2

A. V = Vo + g*Tr
0 = 107 + (-32)Tr
Tr = 3.34 s. = Rise time.
V^2 = Vo^2 + 2g*h.
0 = 107^2 + (-64)h
h = 178.9 Ft. above roof.
ho+h = 76+178.9 = 254.9 Ft. above gnd.
0.5g*Tf^2 = 254.9
16Tf^2 = 254.9
Tf = 4 s. = Fall time.
Tr + Tf = 3.34 + 4 = 7.3 s. = Time to reach gnd.

B. 0.5g*T^2 = 254.9-167
16T^2 = 87.9
2.3 s.