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# a ball is dropped from a height 5m onto a sandy floor and penetrates the sand upto 10cm before coming to rest. Find the retardation of the ball assuming it to be uniform

## 2 Answers

9 years ago

Hi Sudarshana,

Use the equations of motion:

v2-u2 = 2as

Final vel = v = 0

Initial vel = u = √(2g*5) = √(10g)------ {this is obtained by initial vel=0 as droped from a ht, final vel is what u have to determine, height s = 5 mt and accn due to gvt = g, so u2-0=2*g*h or u=√(2gh)}

So 0-10g = 2(a)*(10/100)

So a = -50g.

So retardation is 50*g = 500ms-2.

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

9 years ago

Initial velocity(u) of the ball is 0 m/s.

Let the final velocity of the ball when it just strikes the sand is v m/s.

Applying Newton's laws,we get,

v2-u2=2gs

v2-0=2(10)(5)

v2=100

v=10 m/s.

Now, initial velocity(v) when the ball strikes the sand is 10 m/s

and the final velocity v' when the ball stops inside the sand is 0 m/s.

Here,displacement inside the sand is 10 cm or 0.1 m

and the accn. of the ball inside the sand is a'm/s2.

again,applying Newton's laws,we get,

v'2-v2=2a's

0-100=2(-a)(0.1).........(here the negative sign indicates retardation of the ball,i.e.negative acc. of the ball.)

-100=-0.2a

a=500m/s2

So,the retardation of the ball is 500m/s2.

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