Question

please answer the question quickly

Answer 1

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

begin mathsize 14px style straight t space equals square root of fraction numerator 4 straight h over denominator straight g end fraction end root minus minus minus minus minus minus minus minus minus minus minus minus minus minus left parenthesis 2 right parenthesis end style

From (1) and (2) we get that

begin mathsize 14px style x over v equals square root of fraction numerator 4 h over denominator g end fraction end root x space equals v square root of fraction numerator 4 h over denominator g end fraction end root w e space k n o w space t h a t space x greater or equal than space 0.40 m s o space v square root of fraction numerator 4 h over denominator g end fraction end root space greater or equal than 0.40 m 0.20 space v greater or equal than 0.4 v greater or equal than 2 space m divided by s end style

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