Question

A punter on Westview High School's football team kicks a ball straight up from 5 feet above the ground with an initial upward velocity 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?

Answer 1

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
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