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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).
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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