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Question

Two bodies are thrown vertically upward, with the same initial velocity of $$98\ m/s$$ but $$4\ sec$$ apart. How long after the first one is thrown when they meet ?


A
10 sec
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B
11 sec
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C
12 sec
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D
13 sec
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Solution

The correct option is C $$12\ sec$$
Both the bodies meet each other at t secs after the first one is thrown
Displacement covered by first object in t secs $$s_1=ut-\dfrac{1}{2}gt^2=98t-\dfrac{1}{2}9.8\times t^2$$
Displacement covered by second object in (t-4) secs $$s_2=98(t-4)-\dfrac{1}{2}9.8\times (t-4)^2$$
Since they meet $$s_1=s_2$$
We get $$t=12secs$$


Physics

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