Question

A ball rolls off the top of a staircase with a velocity u. If the height of each step is h, breadth of each step is b, and the ball hits the edge of the nth step, then prove $n = \frac{2hu^{2} }{gb^{2} }$

Answer 1

Answer:

We know that the equation of a trajectory is

y=xtanθ−

2u

2

Cos2θ

gx

2

, where,

y is the y-coordinate of the particle

x is the x-coordinate of the particle

θ is the angle of projection

u is the initial velocity

g is the acceleration due to gravity

In this case we can clearly observe that θ is 0

o

⟹ y=−

2u

2

gx

2

For the ball to hit the n

th

step , it has to travel a vertical distance of nh and nb

∴Upon substituting the values y=−nh and x=nb we get,

−nh=−

2u

2

gn

2

b

2

⟹ n=

gb

2

2u

2

h