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)