Question

A 5kg stone falls from a height of 100m and penetrates 2m in a layer of sand. The time of penetration is ?

Answer 1

answer is 0.29sec

Answer 2

ok...one min..

Answer 3

sure

Answer 4

is it dropped cheche?

Answer 5

It falls from a height ,thats what given..

Answer 6

velocity during approach to sand is 44.72m/s (use formula V=sqrt(2gh)
Vi=44.72,Vf=0
v=u-at
t=(u-v)/a=44.72/a
s=0.5*a*t^2=0.5*(44.72/t)*t^2
4=44.72t
t=4/44.72 = 0.089s,is it to be done like this????

Answer 7

ya tahts exactly wat i got

Answer 8

so i dont get whats wrong???

Answer 9

ahaa!!

now see. the stone has achieved a max velocity (2gh)^(1/2) just before it hits the ground.

and decelerates to 0, while it travels 2m in the sand


so you have initial and final velocity, distance travelled, find deceleration rate. then find the time taken for that decelration!!

Answer 10

rama that wat we did!

Answer 11

LOLLOL

Answer 12

lol

Answer 13

taht mass=5kg data gives us no substantial use

Answer 14

s=0.5*a*t^2=0.5*(44.72/t)*t^2
Sarkar that is not so
S=Vo*t - 1/2 *a* t^2
t=Vo/a
and so
S=Vo^2/2a

Answer 15

yes i thought of that@salinl