Respuesta :
The speed of the stone upon impact with the ground [tex]{V_f}={V_o} (\frac{w-f}{w+f})^{\frac{1}{2}}[/tex]
Given
The initial speed = [tex]{V_o}[/tex]
Explanation
Acceleration
- The state of going quickly or taking place quickly.
- acceleration: the rate at which the speed and direction of a moving object vary over time.
- A point or object going straight ahead is accelerated when it accelerates or decelerates.
Acceleration when ball is moving upward = (w + f)/m
[tex]$\mathrm{V}_{\mathrm{f}}^{2}-\mathrm{V}_{\mathrm{i}}^{2}=2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{s}}$[/tex]
[tex]$=0-\mathrm{V}_{0}^{2}=-2 \mathrm{~s}\left(\frac{\mathrm{w}+\mathrm{f}}{\mathrm{m}}\right)$[/tex] ..... (i)
[tex]$\mathrm{s}=\frac{\mathrm{V}_{0}^{2}}{2\left[\frac{\mathrm{w}+\mathrm{f}}{\mathrm{m}}\right]}$[/tex]
While coming down
[tex]{V_i=0}[/tex]
[tex]$\mathrm{V}_{\mathrm{f}}^{2}-\mathrm{V}_{\mathrm{i}}^{2}=2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{s}} ; \mathrm{a}=(\mathrm{w}+\mathrm{f})$[/tex]
[tex]$\mathrm{V}_{\mathrm{f}}=\sqrt{\frac{2 \times(\mathrm{w}-\mathrm{f}) \mathrm{V}_{0}^{2}}{2(\mathrm{w}+\mathrm{f})}}$[/tex]
[tex]$\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\circ}\left(\frac{\mathrm{w}-\mathrm{f}}{\mathrm{w}+\mathrm{f}}\right)^{1 / 2}$[/tex]
The speed of the stone upon impact with the ground
[tex]$\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\circ}\left(\frac{\mathrm{w}-\mathrm{f}}{\mathrm{w}+\mathrm{f}}\right)^{1 / 2}$[/tex]
Learn more about acceleration here:
https://brainly.com/question/605631
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