Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.)

Question

A baseball is thrown upward from a height of 3 meters with an initial velocity of 8 meters per second. Determine its maximum height. (Round your answer to two decimal places.)

Answer 1

h(t) = -4.9t^2 + 8t + 3

take the derivative and set equal to zero

h'(t) = -9.8t + 8 = 0

max height is when t = 8/9.8 = 4/4.9 = about .816 seconds t

plug that value into the 1st equation above to calculate maximum height'

h(4/4.9) = -4.9(4/4.9^2) + 8(4/4.9) + 3 = -3.2653 +26.1224 +3 = 25.86 meters = max height