tough calc projectile Q w/ incline. HELP

[asourpatchkid]

A ball is hit at an angle \theta to the horizontal, from the base of a hill \phi degrees in incline. The ball strikes the hill at a horizontal distance x measured from the launch point. Determine the value of \theta that will result in the largest possible value of x .

well i started out finding the vertical displacment to be x\tan\phi
then i put that in the basic S_{y} = V_{i}sin\theta t + \frac{1}{2}-9.8t^2 of corse S_{y} = xtan\phi. now i am stumped on what to do. please help!!!

[suffian]

Let y_1 = x tan ph and
Let y_2 = v_0(sin th)t - (1/2)gt^2, x = v_0(cos th)t

note that t can be eliminated from y_2 by putting t = x/[ v_0 cos th ]

[asourpatchkid]

so then i come up with y_2 = x tan \theta - \frac {4.9x^2}{(v_0^2)(cos /theta^2)} if i set that equal to x tan \phi , i can simplify to tan \phi = tan \theta - \frac {4.9x}{(v_0^2)(cos^2 \theta)}

what would i do next??

[suffian]

right, solve the equation you just found for x. Note that this is really a function x(th) that gives you the horz distance x at which the ball hits the hill when thrown at an angle th. Now you can use calculus methods to find the max of this function. be very careful when using inverse functions (like arctan). you might find the following relation very helpful as well, 2 sin th cos th = sin 2 th, use it when you can

[asourpatchkid]

ok i got : x = (v_0/g) (v_0 sin2th - 2y cos th). if i take the derivitive of his equation (velocity) how would i find the derivitve of "y" ??

[suffian]

i'm not sure the equation you got is correct (it might be..), but if you look at your earlier post, you will see an equation with an 'x' but no 'y'. Just solve this for 'x' and you have the function x(th).

btw, because you want to maximize x(th) with respect to theta, make sure to take the derivative with respect to th and notice that all the other variables remain constant in the problem as th changes, so you can treat them as constants in computing the derivative.