Q 42. A staircase contains
three steps each 10 cm high and 20 cm wide (figure). What should be the minimum
horizontal velocity of a ball rolling off the uppermost plane so as to hit
directly the lowest plane?
Answer: To hit directly the lowest plane the ball will have to clear point B. From A to B the ball will move a horizontal distance equal to 40 cm = 0.4 m and the vertical distance equal to 20 cm =0.2 m. Let t be the time taken to reach point be. In this time vertical movement under the gravity has following data,
u=0, h=0.2 m, from the equation h=ut+½gt² , we get,
0.2 = 0+½*9.8 t² → t²=0.4/9.8 = 0.04 → t=0.2
Let v be the horizontal velocity of the ball, then the horizontal distance traveled by the ball = 0.2 v . In order to clear point B this must be at least =0.4,
0.2 v= 0.4 → v=2 m/s. So minimum velocity required to to hit the lowest floor is 2 m/s.
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.
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