Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 10. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?

A. 1.13 s; 30.25 ft

B. 2.25 s; 10 ft

C. 1.13 s; 70.75 ft

D. 1.13 s; 32.5 ft

Answer 1

a. 1.13 s

b. 30.25 ft

Explanation:

Both questions can be answered by putting the equation into vertex form.

h = -16(t^2 -9/4t) +10 . . . . factor the leading coefficient from the t terms

Complete the square by adding the square of half the t coefficient inside parentheses and the opposite of the same amount outside parentheses.

h = -16(t^2 -9/4t +81/64) +10 +81/4

h = -16(t -1.125)^2 + 30.25

This tells us the vertex of the height curve is at (t, h) = (1.125, 30.25), which rounds to ...

• t = 1.13 seconds
• h = 30.25 feet