a stone is dropped from the top of a tower 100 m high.the stone peneterares in the sand on the ground through a distance of 2m. calculate the retardation of the stone
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Answered by
16
Hey.
Here is the answer.
s = 100 m
as, v^2 - u^2 = 2 a s
so, v^2 = 2 × 10 × 100
or, v^2 = 2000
or, v = √2000 m/s
So the stone hits the send with a velocity of √2000 m/s
Now, s = 2 m
let retardation be a
u = √2000 , v = 0
so, v^2 - u^2 = 2as
or, 0 - 2000 = 2×a×2
or, a = - 2000/4 = -500 m/s^2
So, retardation is -500 m/s^2.
Thanks.
Here is the answer.
s = 100 m
as, v^2 - u^2 = 2 a s
so, v^2 = 2 × 10 × 100
or, v^2 = 2000
or, v = √2000 m/s
So the stone hits the send with a velocity of √2000 m/s
Now, s = 2 m
let retardation be a
u = √2000 , v = 0
so, v^2 - u^2 = 2as
or, 0 - 2000 = 2×a×2
or, a = - 2000/4 = -500 m/s^2
So, retardation is -500 m/s^2.
Thanks.
archanaxx:
why you took u=√2000, when firstly you showed v=√2000 by the 3rd equation of motion
as the stone hit sand with v= √2000 but then it penetrated into ground for further 2m so there i considered √2000 as intial velocity and 0 as final velocity as it stops(0) after 2m into the sand.
Answered by
5
from 100m of hight
speed just before striking with ground
= √(2×g×100)
u = √2000
by using formula
v² = u² + 2as
0 = 2000 - 2×a×2
500 = a
retardation = 500m/s²
hope it helps you
speed just before striking with ground
= √(2×g×100)
u = √2000
by using formula
v² = u² + 2as
0 = 2000 - 2×a×2
500 = a
retardation = 500m/s²
hope it helps you
as it comes from hight 100m
speed of any object dropped from rest
u = √2gh
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