Home
Class 11
PHYSICS
A balloon in ascending vertically with a...

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.

Text Solution

Verified by Experts

Let at any time `t=0`, the balloon be position `A`, where its velocity is `u` . At `t=2 s`, it reaches `B`, where its velocity becomes `v`, then
.
`AB=S=uxx2+(1)/(2)a(2)^(2)=2u+2`
Also `v=u +a xx2=u +2`,
First stone: `-S_(1)=uxx3.5 +(1)/(2)(-g)(3.5)^(2)`
Second stone: `S^(2)=vxx1.5+(1)/(2)(-g)(1.5)^(2)`
Required distance between the stones
`x=S_(1)+S-S_(2)`
Solve to get `x=55 m`,
Alterntively: This method can be understood in proper way after studying relative velocity
If we work from the frame of balloon, then the acceleration of each stone w,r.t balloon will be `g+a` after releasing from it. The initial velocity of each stone will be zero w.r.t. balloon.
.
`S_(1)(1)/(2)(g+a)(3.5)^(2), S_(2)=(1)/(2)(g+a) (1.5)^(2), x=S_(1)-S_(2) =55 m`.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Exercise 4.1|17 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Exercise 4.2|12 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Integer|9 Videos

  • GRAVITATION

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos

  • KINEMATICS-2

    CENGAGE PHYSICS|Exercise Exercise Integer|9 Videos

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 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))

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.

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 balloon starts rising from the ground with an acceleration 2 ms^(-2) . After 5 second , a stone is released from the ballooon. Find the time taken by the stone to reach the ground after its release and total height attained by ballooon when stone reaches the ground. Take g==10 ms^(-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)

A balloon reses rest on the ground with constant accleeration 1 m s^(-2) . A stone is dropped when balloon has risen to a height of 39.2 m . Find the time taken by the stone to teach the ground.