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#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 312

I don't know how to do it if the initial velocity isn't known. Can someone help me?

- Thread starter Monoxdifly
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- Thread starter
- #1

- Aug 6, 2015

- 312

I don't know how to do it if the initial velocity isn't known. Can someone help me?

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- #2

\(\displaystyle a=-g\)

And so the velocity $v$ along the vertical axis is:

\(\displaystyle v=-gt+v_0\sin(\alpha)\)

And the height $y$ is:

\(\displaystyle y=-\frac{g}{2}t^2+v_0\sin(\alpha)t=t\left(-\frac{g}{2}t+v_0\sin(\alpha)\right)\)

When the ball reaches the maximum height, we know \(\displaystyle v=0\implies t=\frac{v_0\sin(\alpha)}{g}\) and so:

\(\displaystyle y_{\max}=y\left(\frac{v_0\sin(\alpha)}{g}\right)=\frac{v_0\sin(\alpha)}{g}\left(-\frac{g}{2}\cdot\frac{v_0\sin(\alpha)}{g}+v_0\sin(\alpha)\right)=\frac{v_0^2\sin^2(\alpha)}{2g}\implies v_0\sin(\alpha)=\sqrt{2gy_{\max}}\)

What do we know about $y$ when the ball returns to the ground?

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- #3

- Aug 6, 2015

- 312

y = 0?

Where do I substitute that?

Where do I substitute that?

- Mar 1, 2012

- 980

at max height $v_y = 0$In a soccer match Putu kicked the ball with the elevation angle [FONT=MathJax_Math]α[/FONT] so that it reached the maximum height 10 m. How long must he wait until the ball fall to the ground if g = [FONT=MathJax_Main]10[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main]2[/FONT]?

from the top of the trajectory to the ground ...

$\Delta y = v_y \cdot t - \dfrac{1}{2}gt^2$

$-10 = 0 - 5t^2$

solve for $t$ ...

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- #5

Well, we now have:y = 0?

Where do I substitute that?

\(\displaystyle y=t\left(-\frac{g}{2}t+\sqrt{2gy_{\max}}\right)\)

So, set $y=0$ and find the non-zero root...

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- #6

- Aug 6, 2015

- 312

\(\displaystyle t=\sqrt2\)?

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- #7

The non-zero root is:\(\displaystyle t=\sqrt2\)?

\(\displaystyle t=2\sqrt{\frac{2y_{\max}}{g}}\)

To find the time $t_F$ the ball falls from the apex of its trajectory, we need to subtract the time it took to get there:

\(\displaystyle t_F=2\sqrt{\frac{2y_{\max}}{g}}-\sqrt{\frac{2y_{\max}}{g}}=\sqrt{\frac{2y_{\max}}{g}}\)

We should expect this, as by the symmetry of motion, the ball takes the same anount of time to rise as it does to fall.

Plugging in the data, we find:

\(\displaystyle t_F=\sqrt{\frac{2(10\text{ m})}{10\dfrac{\text{m}}{\text{s}^2}}}=\sqrt{2}\text{ s}\quad\checkmark\)