Suppose the height of a punt (in feet) after t seconds can be modeled by the functionh(t) = 2 + 56t - 16t2a) What is the maximum height of the ball during the punt (5 pts)?b) How long after being kicked will it take the ball to hit the ground (round to the nearesttenth of a second)?

Answer:a) 51 feetb) 3.5 seconds Step-by-step explanation:The y-coordinate of the vertex of the given parabola is what we are looking for.We first need to find the t-coordinate of the vertex.The t-coordinate can be found using -b/(2a).We need to compare -16t^2+56t+2toat^2+bt+c to identify a,b, and c.a=-16b=56c=2We are ready to compute -b/(2a).-b/(2a)=-56/(2*-16)=-56/-32=7/4.The vertex occurs at t=7/4.To find y, we use y=2+56t-16t^2y=2+56(7/4)-16(7/4)^2y=51So the maximum height is 51 feet. Part b)Hitting ground means the height between the ball and the ground is 0.So we need to replace h(t) with 0.0=2+56t-16t^2I'm going to use quadratic formula.a=-16b=56c=2The quadratic formula is:[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex][tex]t=\frac{-56 \pm \sqrt{56^2-4(-16)(2)}}{2(-16)}[/tex]Computing the thing inside the square root and the thing in the denominator using my handy dandy calculator:[tex]t=\frac{-56 \pm \sqrt{3264}}{-32}{/tex]I'm going to do the square root of 3264 now:[tex]t=\frac{-56 \pm 57.131427428}{-32}[/tex].I'm going to compute both[tex]\frac{-56 + 57.131427428}{-32}[/tex] and [tex]-56-57.131427428}{-32}[/tex] using my handy dandy calculator:[tex]-0.035357[/tex] while the other one is [tex]3.535357[/tex]The negative value doesn't make sense for our problem so the answer is approximately 3.5 seconds.