A ball is thrown into the air from a height of 256 feet at time t=0. The function that models this situation is h(t)=-16t^2 + 96t + 256, where t is in seconds and h is the height in feet.A.) what is the height of the ball at 2 seconds?B.) When will the ball reach a height of 144 feet?C.) When will the ball hit the ground?

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]soFor [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 tso[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] thereforethe 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 tso[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] thereforethe solution is the positive value[tex]t=8\ sec[/tex]