Question

A stone weighing 3kg falls from the top of a tower 200 m high and buries itself 2m deep in the sand. Retardation by sand is x × 10^2 m/s. Find x.

Answer 1

Answer:

x=10

Explanation:

velocity gained till ground level:

v²=2gh

v=2×√1000

now for retardation:

distance travelled in sand is 2m

retardation is -a

so V²=u²+2(-a)s

V=0 &u=v=2×√1000

so -u²=2(-a)s

-4000=-2a(2)

a=1000m/s²

Answer 2

Answer:

To find the velocity at the ground level we can use the formulae of newtons laws of motion.

We know that v²=2gh.

v=2×√10*200.

v= 2*√2000.

To calculate the retardation(-a) we get that.

Distance traveled by the stone inside the  sand is 2m.

Hence, V²=u²+2(-a)s.

As final velocity V=0 and u=v=2×√2000.

Now, -u²=2(-a)s.

-8000=-2a(2).

Acceleration a=2000m/s².

Then x=20 .