Question

A boy is standing on a plank which is accelerating
vertically upwards with constant acceleration. If boy
throws a ball vertically up with velocity vo relative
to plank and its time of flight is T, the acceleration
of the plank, is
2V0 +gT
a=
(2) a= Vo+gT
(3) Vogt
(4) Zvo - gT​

Answer 1

Explanation:

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Your answer is : C

Time of fligt , T = 2u/a

T = 2v0/(g+a)

a = (2v0 - gT)/T

Answer 2

ANSWER:

The acceleration of plank $2 \mathbf{v}_{0} / \mathbf{T}-\mathbf{g}$

EXPLANATION:

As the ball is thrown vertically up relative to the upward moving plank, so the acceleration of ball will be uniform and the acceleration of ball will be as follows

acceleration of moving ball relative to plank=-acceleration due to gravity-acceleration of plank

The negative sign before acceleration due to gravity is because the ball is going in opposite direction to the direction of gravity.

Also, we need to find the acceleration of plank, so the above equation will become

acceleration of plank=-acceleration due to gravity-acceleration of moving ball relative to plank

We know the acceleration due to gravity can be represented as g and to determine the acceleration of moving ball relative to plank, we can use one of the equations of motion as follows:

$s=u t+\frac{1}{2} a t^{2}$

Here, s is the displacement of ball relative to plank, u is the velocity of the ball relative to plank, a is the acceleration of ball with respect to plank and t is the time of flight or the total time taken by the ball to complete the action.  

So according to the given problem, s=0 as the ball will return to the same position from where it started in a time period or time of flight of T, u=v0as the velocity of ball with respect to plank is given as v0 and t will be T.

Then,

$s=u t+\frac{1}{2} a t^{2}$

$0=v_{0} T+\frac{1}{2} a T^{2}$

$-v_{0} T=\frac{1}{2} a T^{2}$

$-\frac{2 v_{0}}{T}=a$

If we substitute this “a” in the acceleration of plank equation we will get,

acceleration of plank=-acceleration due to gravity-acceleration of moving ball relative to plank

$acceleration of plank =-g+\frac{2 v_{0}}{T}$

Thus, the acceleration of plank is $2 \mathrm{v}_{0} / \mathrm{T}-\mathrm{g}.$