Projectile Motion on a Staircase

Hi Mohammad, let's solve this classic JEE physics problem where we find the condition for a projected ball to hit the n-th step of a staircase.

We are given that each step has a height a and a width b. The ball is projected horizontally from the edge of the top step with a speed u.

Let's draw a schematic diagram of the staircase and set up our coordinate system at the starting point of the ball.

By placing our origin at the point of projection, the corner of the k-th step is located at a horizontal coordinate k times b, and a vertical coordinate negative k times a.

Now, let us derive the trajectory equation of the projectile.

Along the horizontal direction, there is no acceleration, so the distance covered in time t is simply velocity times time.

In the vertical direction, the ball is under constant acceleration due to gravity, g, downwards.

We can eliminate the time variable from these equations to get the y-coordinate as a function of x.

Next, let's set up the condition for the ball to land on the flat surface of the n-th step.

First, let's examine the condition to clear the corner of the n minus one-th step.

To clear this corner, the trajectory's y-coordinate at this horizontal distance must be strictly greater than the corner's y-coordinate.