Chapter 9.1, Problem 4QC

Suppose the initial conditions in Example 5a are v₀ = 14.7 in/s and s₀ = 49 m. Write the position function s(t), and state its domain. At what time will the stone reach its maximum height? What is the maximum height at that time?

Example 5 Motion In A Gravitational Field

A stone is launched vertically upward with a velocity of v₀ m/s from a point s₀ meters above the ground, where v₀ > 0 and s₀ ≥ 0. Assume the stone is launched at time t = 0 and that s(t) is the position of the store at time t ≥ 0; the positive s-axis points upward with the origin at the ground. By Newton’s Second Law of Motion, assuming no air resistance, the position of the stone is governed by the differential equation s″(t) = −g, where g = 9.8 m/s² is the acceleration due to gravity (in the downward direction)

a. Find the position s(t) of the stone for all times at which the stone is above the ground.