Question

The height of a rock thrown from a cliff can be modeled by the function h(t) = -16t^2+156t+ 306, where t is the

time after the rock is thrown, in seconds.

When does the rock reach its maximum height?

What is the maximum height?

When does the rock hit the ground?

Answer 1

• t = 4.875 s
• h(4.875) = 686.25 ft
• t ≈ 11.424 s

Explanation:

These questions are more easily answered if the equation is written in vertex form. Subtracting the constant and dividing out -16, we get ...

h(t) -306 = -16(t^2 -9.75t)

Adding the square of half the t-coefficient inside parentheses gives ...

h(t) -306 -380.25 = -16(t^2 -9.75t +23.765625)

h(t) = -16(t -4.875)^2 + 686.25

This tells us the maximum height is 686.25 ft at time t = 4.875 seconds.

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When the rock hits the ground, h(t) = 0, so ...

0 = -16(t -4.875)^2 + 686.25

686.25/16 = (t -4.875)^2 . . . . . subtract 686.25, divide by -16

√42.890625 +4.875 = t ≈ 11.424 . . . . take the square root, add 4.875

The rock hits the ground after about 11.424 seconds.