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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

10 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,

10 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.