Question

A particle is projected at an angle of 45 from a point lying 2m from the foot of a wall. It just touches the top of the wall and falls on the ground 4m from it. What is the height of the wall?

Answer 1

projectile Range= R = 2 m + 4m = 6 m

Let g = 10 m/sec²
 let θ = angle of projection = 45 deg

  x = u cosθ t
  y = u sin θ t - 1/2 g t²

=>   y = tan θ x - (g x²) /(2 u² cos² θ)      --- trajectory of the particle
       y  = x - 10 x²/u²
    we know that at x = Range = 6m, y = 0
         =>  0 = 6 - 10 * 36/u²
         =>  u² = 60 units

  equation of projectile = y = x - x²/6

   we know that the wall is at x = 2 meters from the point of projection.
       y  = height of wall = 2 - 10 * 2² /60 = 1.33 meters