A 2.25 kg ball is thrown upward from ground level with an initial velocity of 12.0 m/s. What is the maximum height reached by the ball? Remember to state your givens for initial and final.

Question

Writing down the givens is necessary for both the initial and final. A diagram should also be required.

Answer 1

From kinematics, the equation for the velocity of the ball in m/s as a function of time in seconds is

v(t) = at + v₀

a = -g = -9.8 m/s²

v₀ = 12.0 m/s

v(t) = -9.8t + 12.0

At maximum height, v=0 ⇒

9.8t = 12.0

Solve for t. The result will be in seconds. Then plug the result into the equation for height as a function of time.

h(t) = -4.9t² + 12.0t