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) = -4.9t? + 20t + 65. What is the object's maximum height? The numeric answer only, rounded to the nearest meter.

Answer 1

Answer: 85 meters

Explanation:

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

h(t) = -4.9t² + 20t + 65

a=-4.9 b=20 c=65

`t=(-b)/(2a) = (-(20))/(2(-4.9)) =(-20)/(-9.8)=2`

h(2) = -4.9(2)² + 20(2) + 65

= -19.6 + 40 + 65

= 85.4