Question

From a lift moving upwards with a uniform acceleration $a = 2m{s^{ - 2}}$, a man throws a ball vertically upwards with a velocity \[v = 12m/s\] relative to the lift. The ball comes back to the man after a time of $t$ . Find the value of t in seconds.
A.$3$
B.$2$
C.$4$
D.$6$

Answer 1

Hint: Earth applies a pull on every object called the force of gravity. For an object of mass m, the force of gravity is mg. Now, if this object is on a lift and the lift starts moving upwards with an acceleration a, then the net force on the object will be $m(g + a)$, and when the lift moves downwards the net force experienced by the object is $m(g - a)$.

Complete answer:
Let us first write the information given in the question.
Acceleration of lift $a = 2m/{s^2}$, the ball is thrown vertically upwards with the speed concerning the lift $u = 12m/s$
We have to calculate the times t after which the ball comes back to the man.
Now, we will first consider the lift at rest and write all the parameters concerning the lift.
The acceleration of the ball concerning the lift is given below.
${a_{BL}} = {a_{ball}} - {a_{lift}} = - g - a = - 10 - 2 = - 12m/{s^2}$
Now, let us use the following equation of motion.
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $s$ is the displacement, $t$ is time, $u$ Is initial velocity, and $a$ is the acceleration.
Let us substitute the values in the above equation.
$0 = 12t + \dfrac{1}{2}\left( { - 12} \right){t^2}$
Let us further simplify it by solving the quadratic equation.
$t(12 - 6t) = 0 \Rightarrow t = 0s,t = 2s$

Hence, the correct option is (B). \[2sec\]

Note:
The direct formula to calculate the time for the ball thrown upwards with initial velocity $u$ in a lift is given below.
$t = \dfrac{{2u}}{{g + a}}$
To solve this type of problem, always use the equation of motion and assume one object at rest and calculate all the parameters with respect to it.