Question

# A person in lift which ascends up with acceleration  $$10\ m /s^{2}$$ drops a stone from a height $$10\ m$$. The time of descent is $$(g = 10 m/s^{2})$$

A
1s
B
2s
C
1.5s
D
0.5s

Solution

## The correct option is A $$1 s$$Since, the lift is ascending up against the force of gravity with a = $$10 m/s^2$$, we haveApparent weight of the man $$M(g+a)$$ or the acceleration is $$20 m/s^2$$(also called pseudo acceleration).Now, the stone is dropped from height $$h=10 m$$ and hence time of descent $$t=\sqrt{\dfrac{2h}{g+a}}$$$$t=\sqrt{\dfrac{2 \times 10}{20}}=1 sec$$Physics

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