Suppose that a baseball is thrown upward with an initial velocity of 132 feet per second (90 miles per hour) and it is released 4 feet above the ground. Its height h after t seconds is given by hequalsminus16tsquaredplus132tplus4. After how many seconds does the baseball reach a maximum height?
Accepted Solution
A:
Answer: it will take 8.25 seconds to reach maximum heightStep-by-step explanation:Initial velocity, u of baseball = 132 feet per second. The height of the baseball, h in feet after t seconds is given by the function h = −16t^2 + 132t + 4The given function is a quadratic equation. If values of height attained is plotted against time, the graph will take the shape of a parabola whose vertex corresponds to the maximum height attained by the baseballVertex of the parabola = -b/2aa = - 16b = 132Vertex = - 132/-16× 2 = -132/32Vertex = 4.125The maximum height is 4.125 feetsTo determine the time it will take to reach the maximum height of 4.125 feets, we will substitute h = 4.125 in the equation4.125 = −16t^2 + 132t + 4 −16t^2 + 132t - 0.125 = 0Applying the general formula for quadratic equations,t = [- b ± √b^2 - (4ac)]/2aa = -16b = 132c = -0.125t = [- 132 ± √132^2 - 4(-16 × - 0.125)]/2× -16t = [- 132 ± √17424 - 8)]/-32t = [- 132 ± √17416]/-32t = (-132 ± 132)/-32t = (-132 + 132)/-32 or (-132 -132)/-32t = 0 or -264/-32t = 8.25 seconds