[ASK] How long must he wait until the ball fall to the ground if g = 10m/s^2

[Monoxdifly]

In a soccer match Putu kicked the ball with the elevation angle \(\displaystyle \alpha\) so that it reached the maximum height 10 m. How long must he wait until the ball fall to the ground if g = \(\displaystyle 10m/s^2\)?

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

 

[MarkFL]

Let's let $y$ be our vertical axis of motion, where up is in the positive direction. Neglecting drag, the only force operating on the ball after being kicked is gravity, and so the acceleration $a$ of the ball is:

\(\displaystyle a=-g\)

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?

 

[Monoxdifly]

y = 0?
Where do I substitute that?

 

[skeeter]

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]?

at max height $v_y = 0$

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 $t$ ...

 

[MarkFL]

y = 0?
Where do I substitute that?

Well, we now have:

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

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

 

[Monoxdifly]

\(\displaystyle t=\sqrt2\)?

 

[MarkFL]

\(\displaystyle t=\sqrt2\)?

The non-zero root is:

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

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:

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