Question

a staircase contains three steps each 10 cm high and 20 cm wide what should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane fig given below

Answer 1

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.


Answer:

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.

Answer 2

Answer:2m/s

Explanation:Look first of all we need to understand what question is exactly asking. The Lowermost plane is not the ground but one may consider it as the end of third ie last step of stairs. Now the question goes like this-

height to cover=2xheight of one step.=2x10=20cm=0.2m

Horizontal distance to cover=2xlength of 1 step.=2x20=40cm=0.4m

Total vertical distance=

u(vertical)t +1/2gt^2(considering downward direction as positive)

u(vertical)=0

Total vertical distance=1/2g(t^2)=1/2x10x(t^2) =0.2

(t^2) =2x0.2/10=2x2/100

t=√4/√100=2/10=0.2s

Total horizontal distance=40cm=0.4m

Since it has to hit the Lowermost plane it should cover minimum horizontal distance=0.4m

Thus total horizontal distance should be>=0.4m

Total horizontal distance=u(horizontal)xt

uxt>=0.4m

ux0. 2>=0.4

u>=2m/s

Minimum velocity required to hit the Lowermost plane=2m/s.