Question

A lift ascends with a constant acceleration of 2 m/s2. A man throws a ball vertically upward with a velocity of 12 m/s relative to lift. The ball will come back to man after a time of___ s. (g = 10 m/s2)

Answer 1

Answer:

t=2sec

Explanation:

As it is a non-inertial frame the net acceleration of the particle will be the sum of the acceleration due to gravity and the acceleration of the elevator.

=10+2=12m/s ²

When ball  will fall back into his hands the displacement of the particle will be zero with respect to the man

Also initial velocity of coin, u=20 m/s

Using formula for displacement:-

s=ut+  1/2 at²

0=12t − 1/2 (12t²)

0=12t − 6t²

0=t(12-6t)

0/t=12-6t

0=12-6t

6t=12

t=12/6

t=2seconds

Answer 2

Given:

A lift ascends with a constant acceleration of 2 m/s2. A man throws a ball vertically upward with a velocity of 12 m/s relative to lift.

To find:

Time after which the ball comes back to the man.

Calculation:

Acceleration off the ball relative to the lift is as follows :

$\therefore \: a_{bl} = g + a_{l}$

$= > \: a_{bl} = 10 + 2$

$= > \: a_{bl} = 12 \: m {s}^{ - 2}$

Now , time taken to reach max height be t :

$\therefore \: v = u + at$

$= > \: 0 = 12+ ( - a_{bl})t$

$= > \: 0 = 12 - 12t$

$= > \:t = 1 \: sec$

So, time taken for the ball to come back to the man:

$\therefore \: t_{net} = 2t = 2 \: sec$

So, time taken to come back to man is 2 sec.