Home
Class 11
PHYSICS
A body is thrown up in a lift with a vel...

A body is thrown up in a lift with a velocity `u` relative to the lift, and returns to the lift in time `t`. Show that the lift's upward acceleration is (2u-gt)/t

Text Solution

AI Generated Solution

To solve the problem step by step, we will analyze the motion of the body thrown upwards in the lift and apply the equations of motion. ### Step 1: Understanding the Problem A body is thrown upwards with an initial velocity \( u \) relative to the lift. It returns to the lift after a time \( t \). We need to find the upward acceleration of the lift, denoted as \( a_L \). ### Step 2: Setting Up the Equation In the frame of the lift, we can use the equation of motion: \[ ...
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-1

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 4.4|16 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS ENGLISH|Exercise Subjective|29 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 4.2|12 Videos

  • GRAVITATION

    CENGAGE PHYSICS ENGLISH|Exercise INTEGER_TYPE|1 Videos

  • KINEMATICS-2

    CENGAGE PHYSICS ENGLISH|Exercise Exercise Integer|9 Videos

Similar Questions

Explore conceptually related problems

A body is thrown up in a lift with a velocity 5ms^(-1) relative to the lift and the time of flight is found to be 0.8 s. The acceleration with which the lift is moving up is (g=10ms^(-2))

When a body is thrown up in a lift with a velocity u relative to the lift, the time of flight is found to be t . The acceleration with which the lift is moving up is

A coin is released inside a lift at a height of 2m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 9 m//s^(2) downwards. The time after which the coin will strike with the lift is : (g=10 m//s^(2))

The mass of man when standing on the lift is 60 kg. The weight when the lift is moving upwards with acceration 4.9 ms^(-2) is

A coin is dropped in a lift. It takes time t_(1) to reach the floor when lift is stationary. It takes time t_(2) when lift is moving up with costant acceleration. Then

A man in a lift throws a ball vertically upwards with a velocity v and catches it (i) after t_(1) second when the lift is ascending with an upward acceleration of 'a' and (ii) after t_(2) second when the lift is descending downward with the same acceleration 'a'. THe velocity of the ball v is

A lift can move upward or downward. A light inextensible string fixed from ceiling of lift when a frictionless pulley and tensions in string T_(1) . Two masses of m_(1) and m_(2) are connected with inextensible light string and tension in this string T_(2) as shown in figure. Read the questionbs carefully and answer If m_(1)=m_(2) and m_(1) is moving at a certain instant with velocity v upward with respect to lift and the lift is moving in upward direction with constant acceleration (a lt g) then speed of m_(2) with respect to lift :

An open lift is moving upward with velocity 10 m//s. It has an upward acceleration of 2 m//s^2. A ball is projected upwards with velocity 20 m//s relative to ground. Find (a) time when ball again meets the lift (b) displacement of lift and ball at that instant. (c) distance travelled by the ball upto that instant. Take g=10 m//s^2

A lift is moving up with acceleration a A person inside the ligt throws the ball upwards with a velocity u relativeto hand. a. What is the time of flight of the ball? b. What is the maximum height reached by the ball in the lift?

A body of mass 50 kg is hung by a spring balance in a lift. Calculate the reading of the balance when : The lift is ascending with an acceleration of 2m//s^(2) .