Question

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 32t + 6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

Answer 1

It will reach the maximum in 1 second; the maximum height is 22 feet.

We find the maximum by finding the vertex. We first find the axis of symmetry, whose equation is
x = -b/2a
x = -32/2(-16)
x = -32/-32 = 1

This gives us that it reaches the maximum in 1 second.

To find the y-coordinate, the maximum height, we substitute this into the equation:
h(1) = -16(1²)+32(1)+6
h(1) = -16+32+6 = 22

It reaches a height of 22 feet.