Question

A rocket is launched from the top of a 55-foot cliff with an initial velocity of 138 ft/s.

Answer 1

For this case we have that the equation that describes the height is:

$h = -16t ^ 2 + vt + c$

Where,

v: initial speed

c: initial height

Substituting values we have:

$h = -16t ^ 2 + 138t + 55$

Then, to find the time, we use the quadratic formula:

$t =\frac{-b +/-\sqrt{b^2 - 4ac} }{2a}$

Substituting values we have:

$t =\frac{-138 +/-\sqrt{138^2 - 4(-16)(55)} }{2(-16)}$

Rewriting:

$t =\frac{-138 +/-\sqrt{19044 + 3520} }{-32}$

$t =\frac{-138 +/-\sqrt{22564}}{-32}$

$t =\frac{-138 +/-150.21}{-32}$

We discard the negative root because we want to find the time.

We have then:

$t =\frac{-138 -150.21}{-32}$

$t =\frac{-288.21}{-32} \\t=9$

Answer:

D. $0 = -16t^2 + 138t + 55; 9 s$