Answer:
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)
, 
and 
, therefore the gravitational force between the two objects is