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Question

Two bodies are thrown simultaneously from a tower with same initial velocity $${v_0}$$ one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is.



A
2v0t+12gt2
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B
2v0t
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C
v0t+12gt2
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D
v0t
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Solution

The correct option is B $$2{v_0}t$$
For vertically upward motion,
$$h_1=v_0t-\cfrac{1}{2}gt^2$$
and for vertically downward motion,
$$h_0=v_0t+\cfrac{1}{2}gt^2$$
$$\therefore$$ Total distance covered in $$t$$ sec
$$h=h_1+h_2=2v_0t$$

Physics

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