ANSWER
a) 150ft
b) 3.05s
EXPLANATION.
The quadratic function that models the height of the ball is
[tex]h(t) = - 16.1{t}^{2} + 150[/tex]
The ball was dropped at time t=0.
We plug in t=0 into the given function to get,
[tex]h(0) = - 16.1{(0)}^{2} + 150[/tex]
[tex]h(0) = 150[/tex]
Therefore the ball was dropped from a height of 150 ft.
When the ball hit the ground, then h(t)=0.
This implies that:
[tex]- 16.1{t}^{2} + 150=0[/tex]
[tex]- 16.1{t}^{2} =- 150[/tex]
[tex]{t}^{2} =\frac{- 150}{-16.1}[/tex]
We take square root of both sides,
[tex]{t} =\sqrt{9.317}[/tex]
[tex]{t} =3.05[/tex] to the nearest hundredth.
Therefore the ball hit the ground after approximately 3.05 seconds.
