Question

A ball is thrown into the air with an upward velocity of 12 ft/sec. The equation for the ball's height h

after t seconds is h(t) = -4t2 + 12t + 5
a. Find the maximum height attained by the ball.
b. Find the time at which the ball reaches the ground.

Answer 1

h(t) = -4t² + 12t + 5

a. h'(t) = -8t + 12 = 0

8t = 12

t = 1.5 seconds

h(1.5) = 14 feet

b. 4t² - 12t - 5 = 0

t = (12 + √(12² - 4(4)(-5)))/(2×4)

= (12 + √(144 + 80))/8

= (12 + √224)/8

= (12 + 4√14)/8

= (3 + √14)/2 = about 3.37 seconds

The ball reaches the ground in about

3.37 seconds.