Question

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.912 + 20t + 65. What is the

object's maximum height?

Answer 1

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

a=-4.9 b=20 c=65

h(2) = -4.9(2)² + 20(2) + 65

= -19.6 + 40 + 65

= 85.4

Explanation: