Question

A staircase contains three steps each 10 cm high and 20 cm wide (figure 3-E9). What should be minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane ?Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution
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To hit directly the lowest plane the ball has to cross the point B.
From A to B, the  ball has to travel to horizontal distance equal 40cm =40m
and vertical distance of 20cm=0.2m.

Let the time taken to reach point be t.u=0 m/s
h=0.2m
from the equation, h=ut+1/2gt²

we get
0.2=0+1/2 x9.8 t²
t²=0.4/9.8
t²=0.04
t=0.2
Let v be the horizontal velocity of ball than the horizontal distance travelled by the ball=0.2v
in order to clear the point B it must be equal to 0.40.
2v=0.4
v=2m/s
∴So minimum velocity required to to hit the lowest floor  is 2 m/s.    

Answer 2

* Verified answer.

h= height

w= width of each step

O= uppermost plane

A= lowermost plane

Consider the ball roll from the upmost plane O with horizontal velocity v here the vertical velocity in the y direction will be 0

X=vt

t= vx ...................(1)

The horizontal range of the ball lies between 2wand3w

Horizontal range x of the ball will be greater than 2w.

h=10cm=0.10m

w=20cm=0.20m

i.e,x≥2w

x≥3×0.20

x≥0.40m

The ball travels form plane O to plan A=2h

equation of motion, s=ut+( 21 )gt 2

u=0 for vertical motion

2h=ut+( 21 )gt 2

t= g4h ......................(2)

form (1)and (2)

vx = g4h

x= v g4h

we know x≥0.40m

vg4h ≥0.40m

0.20v≥0.40

v≥2m/s