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A rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s. The height of the rocket above the ground at time t is given by h = –16t2 + 156t + 105. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.


4.9 s

9.8 s

0.6 s

10.4 s

Verified Answer :

Nirina7

h = –16t2 + 156t + 105, when t = t', the time where the rocket hit the ground after it is launched, so 105= –16t'2 + 156t' + 105, so -16t'2 + 156t' = 0 implies t' =0 or -16t'+ 156= 0, equivalent to t'=9.75, so the answer is 
b) 9.8 s

Answer:

The correct option is 4. The rocket will hit the ground after 10.4 s.

Step-by-step explanation:

The given function is

\(h=-16t^2+156t+105\)

Where, h is height of the rocket above the ground at time t.

The rocket will hit the ground, when h=0.

\(0=-16t^2+156t+105\)

Using quadratic formula

\(t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

\(t=\frac{-156\pm \sqrt{(156^2-4(-16)(105)}}{2(-16)}\)

\(t=-0.632,10.382\)

The time can  not negative, therefore after 10.382 second the rocket will hit the ground.

\(10.382\approx 10.4\)

Therefore the correct option is 4. The rocket will hit the ground after 10.4 s.

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