A rocket is launched from atop a 101 foot cliff with an initial velocity of 116 ft/s. \a. Substitute the values into the vertical motion formula h(t) = -16t2 + vt + h0 . Let h(t) = 0

b. How long will the rocket take to hit the ground after it is launched? Round to the nearest tenth of a second.

A. 16t2 -116t -101=0 B. 8.0s

Explanation:

Known parameters from the question:

Height of Cliff,h0 = 101ft

Velocity of rocket ,v= 116ft/s.

1.Substituting the above to the above formula we have;

h(t) = -16t2 + vt + h0

Since h(t) =0, it means the rocket is falling towards the ground, when it gets to the ground when it's at rest the height h(t) = 0m

{Note the rocket is launched from the height of the Cliff so that would be the initial height of the rocket,h0}

2.Substituting into h(t) = -16t2 + vt + h0

We have;

0 = -16t2 + 116t + 101=> 16t2 -116t -101=0

Using formula method for solving quadratic equation we have;

t = -(-116)+_√[(-116)^2 -( 4× 16 ×-101]/ (2× 16)

t = [116 +_(141.1382)]/32

t = (116 -141.1382)/32 or (116 +141.1382)/32

-0.786s or 8.036s

-0.8s or 8.0s to the nearest tenth.

Now time cannot be negative in real life situation hence the time is 8.0s

Note : the general equation of a quadratic equation with variable t is given below;

at2 + bt + c=0

Formula method for quadratic equation is :

t =( -b+_√[(b^2 -( 4× a×c)] ) / (2× a)

the inner core.

good luck: )

klicker