Algebra 2, need help

Question

A ball is tossed up into the air from a height of 50 feet. The height, h, of the ball, in feet, is given by the function h(t)=-16t²+64t+50, where t is the time in seconds. What is the maximum height of the ball, in feet?

Answer 1

Take the derivative of h with respect to t. This looks like a calculus question, as the easiest way to solve it. h'(t)=-32t+64 and set it equal to 0. Solve for t. t=2. -32(2)+64=0. Substitute t=2 in the original height equation h(2)=-16(2)² +64(2) + 50 = -64+128+50=114 feet 114 feet is the maximum height of the ball. The ball reaches its maximum height in 2 seconds.

IF you don't know how to take derivatives, which is the first half of calculus, then graph the height equation, plotting h against t. Plot several points, for t=0 h=50, t=1 h=98, t=2 h=114, t=3 h=98. t=4 h=50 Plot a few more points, substituting t=2.5 h=110 and t=1.5 h=110, into the h(t) equation and narrow down where the curve reaches a maximum. Carefully plot the equation and h=114 is the maximum height. The heights are symmetrical around h=114.

Another method is find the times when the height = 50. We already know t=0 is the first time.

Take the original equation and set it equal to 50. That reduces to -16t²+64t=0

Divide by -16, leaving t²-4t=0 factor giving t(t-4)=0. Set each factor = 0.

t=0 and t-4=0 or t=4. At zero seconds and at 4 seconds the ball is at 50 feet.

Half way in between it reaches the maximum height. Half way between 0 and 4 is 2

The ball reaches a maximum height at t=2. Substitute 2 into the original equation

h(2)=-16(2)²+64(2)+50=114 feet for the maximum height.