Question

An open lift is moving upwards with velocity \[10\text{ }m/s\]. It has an upward acceleration of \[2m/{{s}^{2}}\]. A ball is projected upwards with upwards with velocity \[20\text{ }m/s\] relative to the ground. Find the time when the ball again meets the lift.
A. \[\dfrac{5}{3}s\]
B. \[\dfrac{4}{3}s\]
C. \[\dfrac{3}{4}s\]
D. \[\dfrac{1}{4}s\]

Answer 1

Hint: First find the relative velocity of the ball with respect to the lift by subtracting the lift’s velocity from the ball’s velocity when going upwards. After finding the velocity, find the relative acceleration for the acceleration as well and then use the formula of distance with respect to velocity, acceleration and time to find the time taken for the ball to meet the lift after takeoff.
The formula for the distance is:
\[S=ut+\dfrac{1}{2}a{{t}^{2}}\]
where S is the distance the ball is going upwards with respect to the lift, a is the relative acceleration, u is the relative velocity and t is the time taken for the ball to meet the lift again.

Complete step by step answer:
The time taken when the ball again meets the lift is \[t\].
The speed with which the lift moves upward is \[10\text{ }m/\operatorname{s}\].
The speed with which the ball is projected upwards from the ground is \[20\text{ }m/\operatorname{s}\].
The acceleration of the ball moving upward in the lift is given as \[2\text{ }m/{{s}^{2}}\].
According to the velocities of the ball and the lift given the relative velocity of the ball is the velocity of the lift subtracted from the velocity of the ball i.e. \[20-10=10\text{ }m/\operatorname{s}\].
Now using the kinematic law of one dimension, we place the value of velocity and acceleration in the formula of displacement as:
\[S=ut+\dfrac{1}{2}a{{t}^{2}}\]
With the formula placed we can now sense that both the lift and the ball are moving upward with uniform motion the displacement is taken as zero for the ball.
Placing the displacement as zero, we get the time in terms of velocity and acceleration as:
\[0=ut+\dfrac{1}{2}a{{t}^{2}}\]
\[\Rightarrow -ut=\dfrac{1}{2}a{{t}^{2}}\]
\[\Rightarrow t=\dfrac{-2u}{a}\]
Placing the relative velocity of the ball with respect to that of the lift as \[u=10\text{ }m/s\] and \[a=12\text{ }m/s\]. The acceleration is \[a=12\text{ }m/s\] as both the acceleration are applied in opposite direction that the acceleration due to gravity is downwards while the acceleration of the ball is upwards hence, the relative acceleration is \[10\text{ }m/{{s}^{2}}+2\text{ }m/{{s}^{2}}=12\text{ }m/{{s}^{2}}\].
\[t=\dfrac{-2u}{a}\]
\[\Rightarrow t=\dfrac{-2\times 10}{12}\]
\[\Rightarrow t=\dfrac{5}{3}\sec \] (the time can never be taken as negative)
Therefore, the time when the ball again meets the lift is \[t=\dfrac{5}{3}\sec \].

Note:
In relative velocity or acceleration, when the bodies are going in opposite direction the values tend to add up while going in same direction the values subtract each other therefore, the acceleration adds up as the acceleration of the ball is upwards while the gravity is downwards and the velocity of the ball is upwards while that of the lift is also upwards hence, subtracting both of them.