1. ## projectile

a baseball fielder can throw a ball faster at low angles than high angles. this is modelled by assuming that at an angle a he can throw the baseball with a speed of k√(cosa) . where k is a constant. show the horizontal distance he can throw the ball is given by 2k^2/g *sina*√(cosa)

so i am using the equation ut-.5at^2=0 to find this. so i find what the time is when the vertical distance is zero. k√(cosa)sinat-.5gt^2
so time=2
k√(cosa)sinat /g . i the sub this into the equation distance in the x direction=ut so

k√(cosa)cosa ( 2k√(cosa)sinat /g) is what i get but i dont think this is right

2. ## Re: projectile

My first question is, what, exactly, do you mean by "horizontal distance"? Is that when the ball comes back to the same level as when it was thrown (the height of the thrower's arm) or to the ground? Given that he throws at angle a, with initial speed s, the height, above his arm, after time t, is given by $\displaystyle h= -\frac{g}{2}t^2+ s sin(a)t$ and the horizontal distance by $\displaystyle d= s cos(a) t$. Assuming you only asking for the distance when the ball comes back to the same height (the other option requires that we know the initial height) then we need to find $\displaystyle s cos(a) t$ with t such that $\displaystyle -\frac{g}{2}t^2+ s sin(a) t= 0$. Solve that quadratic equation for t, then use that value of t to find $\displaystyle s cos(a) t$.

3. ## Re: projectile

i think they mean when the ball comes back to the same level.

so when i solve that i get t=2ssina/g and then when i sub this in i get 2s^2cosasina/g sadly this is not right

4. ## Re: projectile

sorry in the opening post you have to prove the distance is 2k^2/g *sinacos^2a and not what i said it was

5. ## Re: projectile

range equation ...

$R = \dfrac{v_0^2 \sin(2\alpha)}{g}$

substitute for $v_0 = k\sqrt{\cos{\alpha}}$

$R = \dfrac{ (k\sqrt{\cos{\alpha}})^2 \sin(2\alpha)}{g}$

$R = \dfrac{ k^2\cos{\alpha} \cdot 2\sin{\alpha}\cos{\alpha}}{g}$

$R = \dfrac{ 2k^2\cos^2{\alpha} \cdot \sin{\alpha}}{g}$

6. ## Re: projectile

Thanks. The next part of the question says by writing it in the form of 2k^2 sina cos^2 a find the max distance he can throw the ball and the angle he throws it at to get the max distance.

I know the max angle is usually when sina=1 or do you solve this using calculus

7. ## Re: projectile

Under normal conditions, max range occurs when $\sin(2\alpha)=1$, but not in this case since the launch angle affects the initial speed.

Assuming $k$ is a constant, find $\dfrac{dR}{d\alpha}$ and maximize.

8. ## Re: projectile

So when I'm differentiating it am i supposed to leave k out of it all together?

9. ## Re: projectile

Originally Posted by markosheehan
So when I'm differentiating it am i supposed to leave k out of it all together?
you have the function ...

$R = \dfrac{2k^2}{g} \left(\cos^2{\alpha} \cdot \sin{\alpha} \right)$

you use the constant multiple rule ... the derivative of a constant times a function = the constant times the function's derivative, i.e. ...

$\dfrac{d}{dx} \left[c \cdot f(x)\right] = c \cdot f'(x)$

so ...

$\dfrac{dR}{d\alpha} = \dfrac{2k^2}{g} \cdot \dfrac{d}{d\alpha}\left(\cos^2{\alpha} \cdot \sin{\alpha} \right)$

10. ## Re: projectile

so i get 2k^2/g (cos^2 a -2sin^2a cosa). how do i maximise this? i cant just let it equal to zero like i would normally do because it has a k in it.

11. ## Re: projectile

Originally Posted by markosheehan
so i get 2k^2/g (cos^2 a -2sin^2a cosa). how do i maximise this? i cant just let it equal to zero like i would normally do because it has a k in it.
$R = \dfrac{2k^2}{g}(\cos^2{\alpha} \cdot \sin{\alpha})$

product rule ...

$\dfrac{dR}{d\alpha} = \dfrac{2k^2}{g}\bigg[\cos^2{\alpha} \cdot \cos{\alpha} + \sin{\alpha} \cdot 2\cos{\alpha} \cdot (-\sin{\alpha})\bigg]$

$\dfrac{dR}{d\alpha} = \dfrac{2k^2}{g}\bigg[\cos^3{\alpha} - 2\sin^2{\alpha} \cdot \cos{\alpha}\bigg]$

$\dfrac{dR}{d\alpha} = \dfrac{2k^2}{g}\bigg[\cos{\alpha}\left(\cos^2{\alpha} - 2\sin^2{\alpha}\right)\bigg]$

$\dfrac{dR}{d\alpha} = \dfrac{2k^2}{g}\bigg[\cos{\alpha}\left(1 - 3\sin^2{\alpha}\right)\bigg] = 0$

you only need to set the trig factors equal to zero because $\dfrac{2k^2}{g} \ne 0$ ...

$\cos{\alpha}=0 \implies \alpha = 90^\circ$ ... that would be a minimum, correct?

$1 - 3\sin^2{\alpha} = 0 \implies \sin{\alpha} = \dfrac{1}{\sqrt{3}} \implies \alpha = \arcsin\left(\dfrac{1}{\sqrt{3}} \right) \approx 35^\circ$

12. ## Re: projectile

thanks. just wondering why do you only set the trig factors equal to zero and not 2k^2/g

13. ## Re: projectile

$g \ne 0$. Do you agree with that?

... at an angle a he can throw the baseball with a speed of k√(cosa) .
If $k^2 = 0 \implies k = 0 \implies v_0 = k\sqrt{\cos{\alpha}} = 0$, which means the initial speed of the ball would be zero. Is that what happens?

So, what can you say about the value of $\dfrac{2k^2}{g}$ ?

14. ## Re: projectile

its not zero. but why does this mean i cant multiply it but the trig function and let it equal to zero.?

15. ## Re: projectile

Originally Posted by markosheehan
its not zero. but why does this mean i cant multiply it but the trig function and let it equal to zero.?
huh? multiply what?