Home
Class 12
PHYSICS
Two balls A and B are placed at the top ...

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t=2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in `ms^(-1)`. (use `g=10 ms^(-2)` ]

A

10

B

15

C

20

D

30

Text Solution

AI Generated Solution

Promotional Banner

Topper's Solved these Questions

  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise Physics - Section B|40 Videos

  • JEE MAINS 2022

    JEE MAINS PREVIOUS YEAR|Exercise PHYSICS- SECTION A |19 Videos

  • JEE MAINS 2021

    JEE MAINS PREVIOUS YEAR|Exercise Physics (Section B )|10 Videos

  • JEE MAINS 2023 JAN ACTUAL PAPER

    JEE MAINS PREVIOUS YEAR|Exercise Question|360 Videos

Similar Questions

Explore conceptually related problems

A ball thrown vertically upwards with an initial velocity of 1.4 m/s returns in 2s. The total dispalcement of the ball is

From the top of a toward of height 60 m , a ball is thrown vertically upward with speed 20 m//s . After some time, the ball strikes the ground. Sketch the v-t and s-t graph.

A ball is dropped from the top of a tower. After 2 s another ball is thrown vertically downwards with a speed of 40 ms^(-1) . After how much time and at what distance below the top of tower the balls meet ?

A ball is thrown vertically upwards from the top of a tower with a velocity of 10m/s. If the ball falls on the ground after 5 seconds, the height of the tower will be? (use g=10m/ s^(2) )

A ball is dropped from the top of a tower. After 2 s another ball is thrown vertically downwards with a speed of 40 m//s . After how much time and at what distance below the top of tower the balls meet?

A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower,It hits the ground after some time with a velocity of 80m/s .The height of the tower is : (g=10m/g^(2))

If a ball is thrown vertically upwards with a velocity of 40m//s , then velocity of the ball after 2s will be (g = 10m//s^(2))

A ball is thown upwards from the ground with an initial speed u. The is at a height of 80 m at two times, the interval being 6s. Find the value of u.