Question

a stone is dropped from the top of a tower of height 100M. the stone penetrates in the sand on the ground through a distance of 2m. calculate the retardation of the stone

Answer 1

aloha user!!
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the stone is being dropped from the top of a tower of hright 100 m. 
which means the potential energy is being converted to kinetic energy.

just before hitting the sand:

potential energy = kinetic enegy 

mgh = $\frac{1}{2} m u^{2}$

u² = $\frac{2mgh}{m}$

u² = 2gh

u² = 2 × 9.8 × 100 

u² = 1960

and we also know that:

v² = u² + 2as

as the retardation is negative the sign will get changed to:

v² = u² - 2as 

we will take the final velocity as 0 because the stone comes to rest after hitting the sand.

0 = 1960 - 2 × a × 2             ( given: the stone penetrates in the sand on the ground through a distance of 2m )

-4a = -1960

a= 490 m/$s^{-2}$

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good life...

Answer 2

Answer:

Explanation:

"U" is initial velocity when it is dropped

"V" is the velocity just above the sand

"V not" is the velocity when it stops in the sand