A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s. A. Substitute the values into the vertical motion formula h=-16t2+vt+c. Let h = 0. B. Use the quadratic formula to find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.The work for the solutions would be great !!

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:
[tex]0=-16 t^{2} +122t+99[/tex]
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:
[tex]-99=-16 t^{2} +122t+99[/tex]
[tex]-16 t^{2} +122+198=0[/tex]

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

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