Home
Class 11
PHYSICS
A ball is dropped from the roof of a tow...

A ball is dropped from the roof of a tower height `h`. The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three seconds. The value of `h` in metre is `(g=10m//s^(2))`

Text Solution

Verified by Experts

Let time of fall be 'n'
`(1)/(2)g (2n -1) = (1)/(2) g(3)^(2) `
`rArr n = 5 sec`.
Height of tower = `(1)/(2) g(5)^(2) = 125` m.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN |Exercise BEGINNER S BOX-7|8 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN |Exercise BEGINNER S BOX-5|5 Videos

  • KINEMATICS

    ALLEN |Exercise EXERCISE-2|89 Videos

  • MISCELLANEOUS

    ALLEN |Exercise Question|1 Videos

Similar Questions

Explore conceptually related problems

A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for :-

A ball dropped from the top of a tower covers a distance 7x in the last second of its journey, where x is the distance covered in the first second. How much time does it take to reach to ground?.

A body is projected vertically up. What is the distance covered in its last second of upward motion? (g=10 m//s^(2))

If a ball is thrown vertically upwards with speed u , the distance covered during the last t second of its ascent is

A particle is dropped from the top of a high building 360 m. The distance travelled by the particle in ninth second is (g=10 m//s^(2))

A particle is dropped from the top of a tower. It covers 40 m in last 2s. Find the height of the tower.

A particle is dropped from the top of a tower. During its motion it covers (9)/(25) part of height of tower in the last 1 second Then find the height of tower.

What would be the distance covered by the particle falling freely in 1s? (g = 10 ms^(-2) )

The displacement of a particle is given by x = (t-2)^(2) where x is in metre and t in second. The distance covered by the particle in first 4 seconds is

A ball is dropped from the top of a building the ball takes 0.5 s to fall past the 3 m length of a window at certain distance from the top of the building. Speed of the ball as it crosses the top edge of the window is (g=10 m//s^(2))