A ball is horizontally projected with a speed v from the top of a plane inclined at an angle 45 ° with the horizontal. How far from the point of projection will the ball strike the plane?

Step-by-step solution

From the above diagram, in ∆ O A B , tan 45 ° = O A A B = y x ⇒ y = x     . . . 1 From equation of motion along vertical direction, the displacement is given by, y = u y T + 1 2 a y T 2 ; here u y = 0 , T is time of flight and a y = g y = 0 + 1 2 g T 2     . . . 2 Along horizontal direction, there is no force or acceleration in this direction, so, x = v × T     . . . 3 From equation 1 , 2 and 3 , 1 2 g T 2 = v × T T = 2 v g Hence, we have x = y = v × 2 v