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) = - 4.9t² + 20t + 65. What is the object's maximum height? the numeric answer only, rounded to the nearest meter.

To answer the question above, first you have to take the derivative of the h (t) function and set it equal to zero, then solve for t to find the time when the object is at its maximum height:

h ' (t) = - 9.8 t + 20

- 9.8 t + 20 = 0

- 9.8 t = - 20

t = ( - 20 / - 9.8) = 2.04 seconds

(I will round that to 2.0 seconds, which is acceptable because of the significant figures used in the question, but I won't go into that.)

So the object reaches max. height at t = 2

Put that value of t into the original h (t) function to find the height at time t = 2:

h (2) = ( - 4.9 * 2) + (20 * 2) + 65

h (2) = ( - 9.8 + 40 + 65) = 95.2 meters.