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