A punter on Westview High School's football team kicks a ball straight up from 5 feet above the ground with an initial upward ve
locity of 45 feet per second. The height of the ball above ground after t seconds is given by the equation h = –16t2 + 45t + 5, where h is the height of the ball in feet and t is the time in seconds since the punt. What is the maximum height of the ball, to the nearest foot?
This can solved graphically, using algebraic manipulation or differential calculus.
Plotting the equation will generate a parabola. The vertex represents the point where the ball will reach the maximum height.
The vertex can be determined by completing the square h = -16t2 + 45t + 5 h - 5 = -16(t2 - 45/16t) h - 5 - 2025/64 = -16(t2 - 45/16t + 2025/1024) (-1/16)(h - 2345/64) = (t - 45/32)^2
The vertex is (45/32,2345/64) or (1.41,36.64)
The maximum height is 36.64 ft
Using calculus, taking the first derivative of the equation and equating to 0 dh/dt = 0 = -32t + 45 t = 45/32 Substituting this value to the equation h = -16(45/32)^2 + 45(45/32) + 5 h = 36.64 ft <span />
