Question

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 !!

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= (-122+ √(27556) )/(-32)` or
`t= (-122- √(27556) )/(-32)`

`t= (-122+166)/(-32)` or
`t= (-122-166)/(-32)`

`t= (-11)/(8)` or
`t=9`

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