MHB [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 $$\alpha$$ so that it reached the maximum height 10 m. How long must he wait until the ball fall to the ground if g = $$10m/s^2$$?

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:

$$a=-g$$

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

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

And the height $y$ is:

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

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

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

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 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=MathJax_Math]m[FONT=MathJax_Main]/[FONT=MathJax_Math]s[FONT=MathJax_Main]2?

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:

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

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

 

[Monoxdifly]

$$t=\sqrt2$$?

 

[MarkFL]

$$t=\sqrt2$$?

The non-zero root is:

$$t=2\sqrt{\frac{2y_{\max}}{g}}$$

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:

$$t_F=2\sqrt{\frac{2y_{\max}}{g}}-\sqrt{\frac{2y_{\max}}{g}}=\sqrt{\frac{2y_{\max}}{g}}$$

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:

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