Question

A ball is thrown upwards into the air with an initial velocity of 11.83 m/s. What is the maximum height that the ball traveled during that time (assuming there is no air resistance).

Answer 1

7.11 meters

Explanation

to solve this we need to use this formula:

$y_{\max }=\frac{(v_f-v_{1)^2}}{2g}$

Step 1

Let

$\begin{gathered} v_f=0\text{ ( when it reaches the ma}\Xi mun\text{ heigth)} \\ v_0=\text{ initial sp}eed=11.83\text{ m/s} \\ g=9.8\text{ }\frac{\text{m}}{s^2} \end{gathered}$

now,replace and calculate

$\begin{gathered} y_{\max }=\frac{(v_f-v_{1)^2}}{2g} \\ y_{\max }=\frac{(-11.83\frac{m}{s})^2}{2\cdot9.8\frac{m}{s^2}} \\ y_{\max }=\frac{139.4989}{19.6}m \\ y_{\max }=7.11\text{ m} \end{gathered}$

therefore, the answer is

7.11 meters

I hope this helps you