A Staircase contains three steps each 10cm high and 20cm wide.What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane.

Let h be the height and w be the width of each step.

Let O be the uppermost plane and A be the lowermost plane.

 

 

 

 

Let the ball roll from the uppermost plane O with a horizontal velocity 'v' .Here the vertical velocity (velocity in the y direction will be 0)

If 't' is the time taken by the ball to reach the lowest plane A and 'x' the horizontal range, then we have:

x = v t  

t = x /v        ---------------------(1)

 

Here the horizontal range of the ball lies between 2w and 3 w, or we can say that the horizontal range (x) of the ball will be greater than 2w.

Given h = 10 cm =0.10 m,

         w = 20 cm =0.20 m

i.e. x ≥ 2w

x ≥ 2×0.20m

x≥ 0.40 m

 

The vertical height, through which the ball travels from the plane O to plane A = 2h

From the equation of motion, s=ut+(1/2) gt² 

Here u =o for vertical motion so:

2h = (1/2)gt² 

 

From (1) and (2)  we get that

 

So the minimum horizontal velocity of a ball rolling off from the uppermost plane(O) so as to hit directly the lowest plane(A) of the staircase = 2 m/s. Further for the ball to roll from the point O to reach the ground level the horizontal velocity should be greater than 2 m/s.