What is the maximum height of the ball?

Question

A ball is thrown upward with an initial velocity of 48 ft./s. The height of the ball above ground after (t) seconds is given by the formula h(t)= -16t^2+48t

What is the maximum height of the ball? How long does it take for the ball to reach it’s maximum height?

Answer 1

The distance that the ball travels is given by the quadratic h(t)=-16t²+48t. Because the number in front of the t² expression is negative, we know that the parabola, when graphing height (h as y coordinate) vs time (t as x coordinate), opens downward and the vertex will be the maximum height. We know that the x value of the vertex will be half way between the 2 points where the graph crosses the x axis. Therefore, we must solve for y=-16x²+48x when y=0:

0=-16x(x-3)--->x=0, x=3

Half way between these 2 points is (0+3)/2=3/2--->1.5s. The max height is when t = 1.5s:

h(1.5)=-16(1.5)²+48(1.5)

h_max=-16(2.25)+48(1.5)

h_max=-36+72

h_max=36

Therefore, the maximum height that the ball will travel is 36ft and the time that it will take to get there is 1.5s.