Question

A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.

Answer 1

0 = –16t2 + 122t + 99; 8.4 s

should be the right answer!!

Answer 2

Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
$0=-16 t^{2} +122t+99$
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
$-99=-16 t^{2} +122t+99$
$-16 t^{2} +122+198=0$

Now we can apply the quadratic formula $t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a}$ where a=-16, b=122, and c=198
$t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}$
$t= \frac{-122+ \sqrt{27556} }{-32}$ or $t= \frac{-122- \sqrt{27556} }{-32}$
$t= \frac{-122+166}{-32}$ or $t= \frac{-122-166}{-32}$
$t= \frac{-11}{8}$ or $t=9$

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.