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

Monoxdifly

Well-known member
Aug 6, 2015
312
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Aug 6, 2015
312
y = 0?
Where do I substitute that?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
980
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Aug 6, 2015
312
\(\displaystyle t=\sqrt2\)?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle t=\sqrt2\)?
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:

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