Question

A rocket is launched from atop a 99-foot cliff with an initial velocity of 122 ft/s.
 a. Substitute the values into the vertical motion formula h = -16t^2 + vt + c. Let h = 0.
 b. Use the quadratic formula find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.

Answer 1

Answer:

8.4 seconds

Step-by-step explanation:

For (a) we have h=0 and t=0    

when we substitute the values we get

$0=-16\cdot 0^2+122\cdot 0+c$

Which will give the value c=0

Now, to find how long the rocket will take to hit the ground after it is launched

we get  $0=-16\cdot t^2+122\cdot t+99$

We will solve the above quadratic equation for t we get:

$16t^2-122t-99=0$

We have formula to solve the quadratic equation:

$t=\frac{-b\pm\sqrt{D}{2a}\text{where}D=-b^2-4ac$

Here, a=16, b=-122 and c=-99

On substituting the values in the formula we get:

$D=(-122)^2-4(16)(-99)=21220$

Now, we will substitute D in the formula to get final value of t

$t=\frac{-(-122)\pm\sqrt{145.67}}{2(16)}$

$t=\frac{122\pm\sqrt{145.67}}{32}$

$t=8.36,-0.73$

We will neglect the negative time

Hence, 8.36 seconds or approximately 8.4 seconds