Question

6) A rock is thrown vertically with an initial velocity of 24.5 m s^-1. What maximum height will the rock reach?

Answer 1

Let's find the time that the rock takes to reach the maximum height:

$$v_y=v_{0}-gt,$$ $$0=v_0-gt,$$ $$t=\dfrac{v_0}{g}.$$

We can find the maximum height reached by the rock from the kinematic equation:

$$y_{max}=v_0t-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the formula, we get:

$$y_{max}=v_0\dfrac{v_0}{g}-\dfrac{1}{2}g(\dfrac{v_0}{g})^2,$$ $$y_{max}=\dfrac{v_0^2}{2g}=\dfrac{(24.5\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}}=30.62\ m.$$