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# A boy throws up a ball in a stationary lift and the ball returns to his hands in $10$ s. Now if the lift starts moving up at a speed of $5$ m/s. The time taken for a ball thrown straight up to return to hand is?

Last updated date: 21st Jul 2024
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Hint:For a body thrown from a point and it returns to the same point then the Displacement of the object is equal to zero. We need to solve the problem keeping in mind the frame, here we can solve the problem in the frame of lift.

Formula to be used:
Second equation of motion: $S\,\,=\,\,u\,t\,+\,\dfrac{1}{2}a\,{{t}^{2}}$
where, $\,u$= initial velocity $a$= acceleration and $t$=time taken.

Given data: time taken by the ball to reach the back to hand, $t=10\,\sec$
speed of lift ${{v}_{L}}\,\,=\,\,5\,\,m/s$
First assign vertically upward direction as positive and vertically downwards as negative.
When a body returns to the same point after moving for some interval of time then its displacement is equal to zero. Hence here when the lift is at rest and the ball returns to hands, in this case displacement of ball is equals to zero.
Applying second equation of motion:
$S\,\,=\,\,u\,t\,+\,\dfrac{1}{2}a\,{{t}^{2}}$
Here, $S\,\,=\,\,0$,$a\,\,=\,\,g\,=\,\,-10\,m/{{s}^{2}}$and $t=10\,\,\sec$

On substituting these terms in the equation we get:
$0\,=\,\,\,u\,(10)\,\,+\,\dfrac{1}{2}\,\,\times \,(-10)\times {{(10)}^{2}}$
$\Rightarrow \,\,0\,\,\,=\,\,\,u\,(10)\,\,+\,\dfrac{1}{2}\,\,\times \,(-10)\times {{(10)}^{2}}$
$\Rightarrow \,\,u\times 10\,\,\,=\,\,\,\dfrac{1}{2}\times 10\times 100$
$\therefore \,\,\,\,u=50\,m/s$ = initial velocity of projection.
Displacement of ball, ${{S}_{B}}\,\,\,=\,\,\,50t\,\,-\,\,\frac{1}{2}\times 10\times {{t}^{2}}$
Displacement of hand = displacement of lift, ${{S}_{L}}=5t$
For ball to return back to hand when ball is thrown in moving lift

Displacement of hand = displacement of ball i.e. ${{S}_{B}}\,=\,{{S}_{L}}$
$\Rightarrow \,\,50t-\dfrac{1}{2}\times 10\times {{t}^{2}}\,\,=\,\,\,5t$
$\Rightarrow \,\,45t=\,\,5{{t}^{2}} \\$
This equals :
${{t}^{2}}=\,\,9t$
$\Rightarrow \,\,{{t}^{2}}\,-\,9t\,\,=\,0 \\$
On factoring the equation, we obtain:
$t(t\,-\,9)\,\,=\,0$
$\therefore \,\,t\,=\,0$ or $t\,=\,9$
Here we are also getting $t\,=\,0$ because at $t\,=\,0$ the ball is in the hand.

Hence after $t\,=\,9\,\,\sec$ the ball will return to his hand.

Note:After solving an equation we might end up in getting more than one root of the equation, in that case the given situation must be analysed and extra roots should be rejected. as in this case that obviously time cannot be zero.