I have encountered a tough question that I certainly need help of. Any help is greatly appreciated.

1) A boy throws a stone to a building a horizontal distance x away through a window of height h above the launching point. The initial (launching) speed is v. "a" is the angle of the initial velocity with the horizontal. Using the projectile motion equations, prove that:

(tan a)^2 - (2v^2 / gx) (tan a) + [1 + (2 v^2 h / g x^2) ] = 0, where g=9.8 m/s^2.

[I am having terrible trouble with this question. Can someone please give me some hints on how to prove this? How many trig identities need to be used?]

2) By solving for tan a in the above equation, find a condition for h such that there will be only 1 angle which will allow the stone to hit the window.

[I used the quadratic formula and try to solve for tan a, but the expression within the square root is crazy, and I can't simplify anything. What should I do?]

Start with the basic setup:

Angle of velocity to horizontal = a
Split velocity into Vx and Vy (velocity along x and y lines respectively).
From basic pythag - V^2 = Vx^2 + Vy^2
Also, Vy/Vx = tan(a)

Projectile equations:
x = Vx t
y = Vy t - 1/2 g t^2

So h = Vy T - 1/2 g T^2 where T = time to reach height h

At this same time, x = Vx T

So T = x / Vx = x tan(a) / Vy

Substitute that into the equation for h:
h = Vy x tan(a) / Vy - 1/2 g (x tan(a) / Vy)^2

We can simplify to: h = x tan(a) - 1/2 g (x tan(a) / Vy)^2

Now we also know that Vy = V sin(a)

So: h = x tan(a) - 1/2 g x^2 tan(a)^2 / (V^2 sin(a)^2)

tan(a)^2 / sin(a)^2 = 1/cos(a)^2 = sec(a)^2
sec(a)^2 = 1 - tan(a)^2

So, substituting all that in, we get:

h = x tan(a) - [g x^2 / 2 V^2](1 + tan(a)^2)

This then reduces, eventually to the equation you're trying to prove.


For the second part...
Simplify the expressions in the above equation to something like:
Z^2 - pZ + (1+hp) = 0
where Z = tan(a) and p = 2 V^2 / gx^2

Solving for Z using the [-b +/- sqrt(b^2 - 4ac)]/2a will only give one result where b^2 - 4ac = 0

In this instance, b = p, a = 1 and c = (1+hp).

So b^2 - 4ac = p^2 - 4 - 4hp

Solve to zero... - is there any value of h which will allow that equation to equal zero? Remember that p is 2V^2/gx^2, so any condition of h that will satisfy this equation = 0 will most likely be related to V and x....

So... merely by rearranging it you get h = (p^2 - 4) / 4p

Thanks a lot for your detailed explanation...