Home
Class 11
PHYSICS
A ballon is ascending vertically with an...

A ballon is ascending vertically with an acceleration fo `0.2m//s^(2)`. Two stones are dropped from it at an interval of `2sec`. Find the distance between them `1.5sec` after the second stone is released. (use `g =9.8 m//s^(2))`

Text Solution

Verified by Experts

The correct Answer is:
`50m`
from the frame attached to ballon
`h_(1) = (1)/(2) (g+a) ((7)/(2))^(2)` & `h_(2) = (1)/(2) (g+a) ((3)/(2))^(2)`
separation between
`(h_(2) -h_(1)) = (1)/(2) xx 10 ((49)/(4) - (9)/(4)) = 50 m`
Promotional Banner

Similar Questions

Explore conceptually related problems

A balloon a ascending vertically with an acceleration of 0.4m//s^(2) . Two stones are dropped from it at an interval of 2 sec. Find the distance between them 1.5sec. after the second stone is released. (g=10m//sec^(2)) .

A balloon in ascending vertically with an acceleration of 1 ms^(-2) . Two stones are dropped from it at an interval of 2 s . Find the distance berween them 1.5 s after the second stone is released.

A balloon is ascending vertically with an acceelration of 0.2 ms^(-2) .Two stones are dropped from it at an interval fo 2 s , the distance between then when the second stone dropped is (tanke g=9.i8 ma^(-2).

A balloon is rising vertically upwards with uniform acceleration 15.7m/ s^(2) .A stone is dropped from it. After 4s another stone is dropped from it.Find the distance between the two stones 6 second after the second stone is dropped.

Two balls are released from the same height at an interval of 2s. When will the separation between the balls be 20m after the first ball is released? Take g=10m//s^(2) .

A stone is let to fall from a balloon ascending with an acceleration f . Aftre time t . A second stonr is dropped. Prove that the distance between the stones after time t' since the second stone is dropped, is (1)/(2) (f+g)t(t+2t') .

A stone is dropped from rest from the top of a tower 19.6m high. The distance travelled during the last second of its fall is (given g=9.8m//s^(2) ):

A stone is let to fall from a balloon ascending with acceleration 4(m)/(s^(2)) . After time t=2s , a second stone is dropped. If the distance between the stones after time (t^(')=3s) since the second stone is dropped is d, find (d)/(28)

balloon is rising vertically up with a velocity of 29 m/s. A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is (g = 9.8 m/s^2)