Question

1. a ball is thrown vertically upward with an initial velocity of 29.4 meters per second. what is the maximum height reached by the ball?

Explanation:

Step1: Identify the relevant kinematic - equation

The kinematic equation $v^{2}=v_{0}^{2}-2gh$ is used. At the maximum - height, the final velocity $v = 0$. The initial velocity $v_{0}=29.4\ m/s$ and the acceleration due to gravity $g = 9.8\ m/s^{2}$.

Step2: Rearrange the equation to solve for $h$

Starting from $v^{2}=v_{0}^{2}-2gh$, when $v = 0$, we can rewrite it as $h=\frac{v_{0}^{2}}{2g}$.

Step3: Substitute the values

Substitute $v_{0}=29.4\ m/s$ and $g = 9.8\ m/s^{2}$ into the formula $h=\frac{v_{0}^{2}}{2g}$. So, $h=\frac{(29.4)^{2}}{2\times9.8}=\frac{864.36}{19.6}=44.1\ m$.

Answer:

44.1 meters