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

In summary, Putu kicks the ball with an elevation angle α so that it reaches the maximum height 10 m. He must wait until the ball falls to the ground before he can stop it. The ball takes 2.5 s to fall from the apex of its trajectory.
  • #1
Monoxdifly
MHB
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0
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?
 
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  • #2
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?
 
  • #3
y = 0?
Where do I substitute that?
 
  • #4
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 $t$ ...
 
  • #5
Monoxdifly said:
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...
 
  • #6
\(\displaystyle t=\sqrt2\)?
 
  • #7
Monoxdifly said:
\(\displaystyle t=\sqrt2\)?

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:

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

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

1. How is the value of g determined?

The value of g, also known as the acceleration due to gravity, is determined by the mass and distance between two objects. It is typically calculated using the formula g = G(m1 + m2)/d^2, where G is the universal gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them.

2. Is the value of g constant for all objects?

No, the value of g can vary depending on the mass and distance of the objects involved. For example, the value of g on Earth is approximately 10m/s^2, but it may be different on other planets or in outer space.

3. How does g affect the speed of an object?

The value of g determines the rate at which an object will accelerate towards the ground. The higher the value of g, the faster the object will fall towards the ground.

4. Can g ever be negative?

No, g cannot be negative. It is always a positive value, as it represents the acceleration due to the gravitational force pulling an object towards the ground.

5. How long does it take for an object to fall to the ground with a value of g = 10m/s^2?

The time it takes for an object to fall to the ground with a value of g = 10m/s^2 can be calculated using the formula t = √(2h/g), where h is the height of the object. For example, if the height is 10 meters, it would take approximately 1.42 seconds for the object to fall to the ground.

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