Question

An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40?

Answer 1

The maximum height is determined by using the given equation as written below.
h(t) = -16t² + 45t + 40

We derive the equation in terms of t and equate the differential to zero.
dh(t) = -32t + 45
0 = -35t + 45

Transpose the constant to the other side of the equation.
35t = 45
Divide the equation by 35, giving us the value of t which is equal to 9/7.

This means that the maximum height is attained after 9/7 seconds.

Substitute the computed time to the equation for height.

h(9/7) = -16(9/7)² + 45(97) + 40

The value of h is equal to 71.41 ft.

ANSWER: 71.41 ft