Home
Class 11
PHYSICS
A ball projected upwards from the foot o...

A ball projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of `6 s` and the ball reaches the ground after `12 s`. The height of the tower is `(g = 10 m//s^2)` :

A

`120m`

B

`135m `

C

`175m`

D

`80m`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS 1

    DC PANDEY|Exercise MCQ_TYPE|27 Videos

  • KINEMATICS 1

    DC PANDEY|Exercise COMPREHENSION_TYPE|19 Videos

  • KINEMATICS

    DC PANDEY|Exercise INTEGER_TYPE|11 Videos

  • LAWS OF MOTION

    DC PANDEY|Exercise Medical entrances gallery|39 Videos

Similar Questions

Explore conceptually related problems

A stone released with zero velocity from the top of a water, reaches the ground in 4s. The height of the tower is (g = 10 m//s^(2))

A stone dropped from the top of a tower reaches the ground in 3 s. The height of the tower is

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 thrown upwards with speed v from the top of a tower and it reaches the ground with speed 3v .The height of the tower is

A ball is thrown upward with speed 10 m/s from the top to the tower reaches the ground with a speed 20 m/s. The height of the tower is [Take g=10m//s^(2) ]

A ball is thrown upwards with speed ‘v’ from the top of a tower and it reaches the ground with a speed 3v, then the height of the tower is

A ball rleased from the tope of a tower travels (11) /( 36) of the height of the tower in the last second of its journey. The height of the tower is (g= 10 ms^(2)).

Two balls are projected simultaneously with the same speed from the top of a tower, one vertically upwards and the other vertically downwards. They reach the ground in 9 s and 4 s, respectively. The height of the tower is (g=10 m//s^(2))