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A stone weighing 3 kg falls from the top of a tower 100m high and buries itself 2m deep in the sand. the time of penetration is

Here,

m = 3kg

h = 100

g = 9.8 ms-2

s = 2 m

Let the velocity with which it reaches ground = u

Now when the stone reaches the foot of the tower its entire PE is converted into KE

So PE = KE

mgh = ½mu2

=> u2 = 2gh  

=> u = √(2gh)

When it penetrates the sand it experiences deceleration a, it stops the ball after 2 m.

Final velocity of the stone v = 0

v2 = u2 + 2as

=> 02 = (√(2gh))2 + 2as

=> 0 = 2gh + 2as

=> a = -gh/s

Again, let t = time taken to penetrate

v = u – at

=> t = (v – u)/a

=> t = {0 - √(2gh)}/(-gh/s)

=>  t = s√(2/gh)

Putting the values

=> t =  2×√(2/(9.8×100))

=> t = 0.09 s

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Solution of:

A stone weighing 2 kg falls from the top of a tower 40 meter high and burries itself 1m deep in sand. The time of penetration is? (take g =9.8ms -1 )

Let v be the speed with which the stone hits the ground. It falls through a height 40 m.

Using, v 2 = u 2 + 2as

= v 2 = 0 + 2 × 9.8 × 40

= v = 28 m/s

Now, the stone penetrates 1 m into sand and comes to rest. Its acceleration is say a and time to stop is say t.

Using, v 2 = u 2 + 2as

= 0 = 28 2 + 2 × a × 1

= a = -392 m/s 2

Since, v = u + at

= 0 = 28 392t

= t = 0.07 s

Solution of:

Three different object of masses m 1 ,m 2 ,m 3 are allowed to fall from rest from the same point along three different frictionless path, the speed of the three object on reaching the ground will be in ratio of?

If the balls falls through the same height, whatever be the frictionless path, their velocity on reaching the ground will be equal. This is according to the law of conservation of energy.

If they fall from a point at a height h from the ground, the potential energy at the point is, PE = mgh

The kinetic energy on reaching the ground is, KE = ½ mv 2

Since, PE = KE

= mgh = ½ mv 2

= v = √(2gh)

The equation is independent of mass.

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