# An elevator is going up with an upward acceleration of 1 m/s^2. At the instant when its velocity...

## Question:

An elevator is going up with an upward acceleration of 1 m/s{eq}^2 {/eq}. At the instant when its velocity is 2 m/s, a stone is projected upward from its floor with a speed of 2m/s relative to the elevator, at an elevation of 30 degrees.

(1) Calculate the time taken by the stone to return to the floor.

(2) If the elevator was moving with a downward acceleration = g, how would the motion be altered?

## Accelerated Frame of Reference

In an inertial frame of reference, Newton's laws will give the correct dynamics. But in an accelerated frame we need to take into account the fictitious forces before applying Newton's laws. If the acceleration of the frame is {eq}\displaystyle {a} {/eq} then a mass {eq}\displaystyle {m} {/eq} will experience a fictitious force {eq}\displaystyle {-ma} {/eq}. Apparently this has no source in the frame. Hence the name fictitious. Once the fictitious force is included then the rest is routine just as in an inertial frame.

## Answer and Explanation:

Here it is given that the stone is projected with an angle of projection of 30 degrees at a speed of 2m/s relative to the elevator frame of reference. Since this is an accelerated frame, the forces acting on the stone are {eq}\displaystyle {mg} {/eq} vertically down and the fictitious force of magnitude {eq}\displaystyle {ma=m\times 1=m} {/eq}, again vertically down since the elevator is accelerating upwards at {eq}\displaystyle {a=1\ m/s^2} {/eq}.

Thus the total force acting on the stone in the elevator frame is,

{eq}\displaystyle {F=mg+ma=m\times 10.8} {/eq}.

Hence the acceleration of the stone is,

{eq}\displaystyle {a'=10.8\ m/s^2} {/eq} vertically down.

Since the stone is projected at 30 degrees with a speed of 2 m/s, the vertical component of its velocity is {eq}\displaystyle { u_x=2 \sin30=1\ m/s} {/eq}.

By the time the stone gets back to the floor its vertical component should become {eq}\displaystyle {v_x=-1\ m/s} {/eq}.

Hence using the velocity -displacement relation,

{eq}\displaystyle {v_x=u_x+a't} {/eq},

We get,

{eq}\displaystyle { -1=1-10.8t} {/eq}.

That is,

{eq}\displaystyle { t=\frac{2}{10.8}=0.185\ s} {/eq}.

NOTE: We have used the convention that upward directed quantities are positive while the downward directed ones are negative.

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Chapter 4 / Lesson 12