Question

What is the angle that allows you to throw the ball the farthest distance given that you are standing a height h above the target? Hints: 1. you can simplify the math recognizing that h = x_ftan ($\varphi$) 2. also you can simplify your answer using cot $\alpha$ = tan( ($\pi$/2) - $\alpha$ )

Answer 1

To solve this question first of all we will find the trajectory of projectile then we apply the condition of maxima by calculus and then using the equations so obtain we elliminate mqximum range Rmax and get the angle..

Suppose projectile is thrown at an angle 'theta'. Let at any time 't' it reached to a point p(x, y). horizontal displacement at time 't' - x = u cos theta times t = t = x/u cos theta ------ (1) for vertical displacement y (from ground) - y = h + h' = h + u sin theta xt minus 1/2 gt^2 ------ (2) Put value of 't' from (1) into (2) - y = h + sin theta times x/ cos theta minus 1/2 g times x^2/u^2 cos^2 theta = y = h + x tan theta minus g/2 u^2 sec^2 theta x^2 ------ (3) Above equation represents the trajectory of projectile. For point A (i.e. the point when projectile hits the target) x = R & y = 0 put this in equ^n (3) \r\n 0 = h + R tan theta mins g/2 u^2 sec^2 theta R^2 ------- (4)