Home
Class 12
PHYSICS
A ball is dropped form the top of a 100...

A ball is dropped form the top of a 100 m high tower on a plant. In the last `(1)/(2)` s before hitting the ground , it covers a distance of 19m . Acceleration due ot gravity (in `ms^(-2)`) near the surface on the planet is ______.

Text Solution

Verified by Experts

The correct Answer is:
79
Time to travel 121 m is t sec
Time to travel 200 m is t `+1/2` sec.
`121 = 1/2 xx a xx t^2 implies t=11 sqrt(2/a)`
`200 = 1/2 xx a xx (t+1/2)^2 implies t+1/2 = sqrt((400)/(a))`
`a=79 m//s^2`
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST 7

    VMC MODULES ENGLISH|Exercise PHYSICS (SECTION 2)|5 Videos

  • MOCK TEST 6

    VMC MODULES ENGLISH|Exercise PHYSICS (SECTION 2)|5 Videos

  • MOCK TEST 8

    VMC MODULES ENGLISH|Exercise PHYSICS (SECTION 2)|5 Videos

Similar Questions

Explore conceptually related problems

A ball is dropped from the top of a 200 m high tower on a planet. In the last 1/2 s before hitting the ground, it covers a distance of 79 m. Acceleration due to gravity (in ms–2) near the surface on that planet is ______.

A ball is dropped from the top of a tower. In the last second of motion it covers (5)/(9) th of tower height before hitting the ground. Acceleration due to gravity = 10 (in ms^(–2)) . Find height of tower (in m):

A ball is dropped from the top of a tower. In the last second of motion it covers (5)/(9) th of tower height before hitting the ground. Acceleration due to gravity = 10 (in ms^(–2)) . Find height of tower (in m):

A particle is dropped from height h = 100 m , from surface of a planet. If in last 1/2 sec of its journey it covers 19 m. Then value of acceleration due to gravity that planet is :

A particle is dropped from height h = 100 m , from surface of a planet. If in last 1/2 sec of its journey it covers 19 m. Then value of acceleration due to gravity that planet is :

A stone is dropped from the top of a tower of height 100 m. The stone penetrates in the sand on the ground through a distance of 2m. Calculate the retardation of the stone.

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

A ball is dropped from a height. If it takes 0.200 s to cross thelast 6.00 m before hitting the ground, find the height from which it was dropped. Take g=10 m/s^2 .

The mass of a planet and its diameter are three times those of earth's. Then the acceleration due to gravity on the surface of the planet is : (g =9.8 ms^(-2))

A ball is dropped from the top of a tower of herght (h). It covers a destance of h // 2 in the last second of its motion. How long does the ball remain in air?