Answer:
Part A) The height of the ball at 2 seconds is [tex]384\ ft[/tex]
Part B) [tex]t=7\ sec[/tex]
Part C) [tex]t=8\ sec[/tex]
Step-by-step explanation:
Part A) what is the height of the ball at 2 seconds?
we have
[tex]h(t)=-16t^{2} +96t+256[/tex]
so
For [tex]t=2\ sec[/tex]
Substitute the value of t in the equation and solve for h
[tex]h(2)=-16(2^{2}) +96(2)+256[/tex]
[tex]h(2)=384\ ft[/tex]
Part B) When will the ball reach a height of 144 feet?
Substitute the value of [tex]h(t)=144\ ft[/tex] in the equation and solve for t
so
[tex]144=-16t^{2} +96t+256[/tex]
[tex]-16t^{2} +96t+112=0[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-16t^{2} +96t+112=0[/tex]
so
[tex]a=-16\\b=96\\c=112[/tex]
substitute in the formula
[tex]t=\frac{-96(+/-)\sqrt{96^{2}-4(-16)(112)}} {2(-16)}[/tex]
[tex]t=\frac{-96(+/-)\sqrt{16,384}} {-32}[/tex]
[tex]t=\frac{-96(+/-)128} {-32}[/tex]
[tex]t=\frac{-96(+)128} {-32}=-1[/tex]
[tex]t=\frac{-96(-)128} {-32}=7[/tex]
therefore
the solution is the positive value
[tex]t=7\ sec[/tex]
Part C) When will the ball hit the ground?
Substitute the value of [tex]h(t)=0\ ft[/tex] in the equation and solve for t
so
[tex]0=-16t^{2} +96t+256[/tex]
[tex]-16t^{2} +96t+256=0[/tex]
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-16t^{2} +96t+256=0[/tex]
so
[tex]a=-16\\b=96\\c=256[/tex]
substitute in the formula
[tex]t=\frac{-96(+/-)\sqrt{96^{2}-4(-16)(256)}} {2(-16)}[/tex]
[tex]t=\frac{-96(+/-)\sqrt{25,600}} {-32}[/tex]
[tex]t=\frac{-96(+/-)160} {-32}[/tex]
[tex]t=\frac{-96(+)160} {-32}=-2[/tex]
[tex]t=\frac{-96(-)160} {-32}=8[/tex]
therefore
the solution is the positive value
[tex]t=8\ sec[/tex]
