h(t)=-16ft^2+40t+6

Question

A ball is thrown into the air with an initial upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function h(t)=-16ft^2+40t+6

In how many seconds will the ball reach the height of 30 feet?

In how many seconds will the ball reach its maximum height?

What is the ball's maximum height?

Answer 1

-16t^2 +40t + 6 = 30

-16t^2 +40t -24 = 0

4t^2 -10t +6 =0

(4t-6)(t-1) = 0

t = 6/4 = 3/2, t = 1. It reaches 30 feet in 1 second and again on the way down in 1.5 seconds

take the derivative and set = 0 to find max height time

-32t +40 = 0

t = 40/32 = 10/8 = 5/4 = 1.25 seconds for max height

plug back into the original h(t) equation to solve for max height

h(5/4) = -16(5/4)^2 + 40(5/4) + 6 = -25 +50 + 6 = 31 feet