1. ## The cricket-ball problem.

Have been tearing my hair out over this for hours...
If you were to hit a cricket-ball a distance of x metres and to a height of x metres, what would be the angle between the ball and the ground upon hitting it. You're expected to assume that the ball doesn't bounce and that g=9.8. Any help, even a point in the right direction, would be greatly appreciated.

2. ## Re: The cricket-ball problem.

Originally Posted by Metricprocess
Have been tearing my hair out over this for hours...
If you were to hit a cricket-ball a distance of x metres and to a height of x metres, what would be the angle between the ball and the ground upon hitting it. You're expected to assume that the ball doesn't bounce and that g=9.8. Any help, even a point in the right direction, would be greatly appreciated.
The trajectory would be a parabola...

3. ## Re: The cricket-ball problem.

I realise that, I've tried every imaginable suvat equation for every possible direction of motion but I'm still no closer. Is there any way to do it by differentiation or something more graphical than plugging in equations?

4. ## Re: The cricket-ball problem.

The ball reaches a maximum height of $\displaystyle x$ metres, and travels $\displaystyle x$ metres horizontally. So if you use the point where the ball is hit as the origin, the turning point is at $\displaystyle \left(\frac{x}{2}, x\right)$.

So substitute these into the equation $\displaystyle Y = a\left(X - h\right)^2 + k$, where $\displaystyle \left(h, k\right)$ is the turning point, and solve for $\displaystyle a$, then you have the equation of your parabola.

Now, the angle made at the origin will be the same as the angle to the tangent at the origin... You should know that for a straight line $\displaystyle m = \tan{\theta}$...

5. ## Re: The cricket-ball problem.

So I've found a in terms of x, but I'm not sure how to use this equation. Do I differentiate it? I realise this is probably embarrassingly easy but please bear with me, thanks for your help so far.

6. ## Re: The cricket-ball problem.

Wait, is it approximately 76 degrees?

7. ## Re: The cricket-ball problem.

Originally Posted by Metricprocess
So I've found a in terms of x, but I'm not sure how to use this equation. Do I differentiate it? I realise this is probably embarrassingly easy but please bear with me, thanks for your help so far.
Yes, differentiate the function to get $\displaystyle \frac{dY}{dX}$ and evaluate it at $\displaystyle X = 0$. This gives you the value of $\displaystyle m$...

8. ## Re: The cricket-ball problem.

Originally Posted by Prove It
Yes, differentiate the function to get $\displaystyle \frac{dY}{dX}$ and evaluate it at $\displaystyle X = 0$. This gives you the value of $\displaystyle m$...
Wait, just to check I'm doing this right, isn't it da/dx seeing as those two are the only variables in my equation.

9. ## Re: The cricket-ball problem.

Originally Posted by Metricprocess
If you were to hit a cricket-ball a distance of x metres and to a height of x metres, what would be the angle between the ball and the ground upon hitting it. You're expected to assume that the ball doesn't bounce and that g=9.8.
range ...

$x = \frac{2v^2 \sin{\theta}\cos{\theta}}{g}$

height ...

$x = \frac{v^2 \sin^2{\theta}}{2g}$

$\frac{2v^2 \sin{\theta}\cos{\theta}}{g} = \frac{v^2 \sin^2{\theta}}{2g}$

$2\cos{\theta} = \frac{\sin{\theta}}{2}$

$4 = \frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}$

$\theta = \arctan(4) \approx 76^\circ$

10. ## Re: The cricket-ball problem.

Originally Posted by Metricprocess
Wait, just to check I'm doing this right, isn't it da/dx seeing as those two are the only variables in my equation.
You should have a in terms of x, and x is treated like a constant, since it's a distance.

Y and X are the variables.

11. ## Re: The cricket-ball problem.

Many thanks, I somehow managed to do the whole question in 3 lines of working 'just by looking at it' but unsurprisingly that's not good enough. Thanks a million