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Question

# An object thrown vertically up from the ground passes the height 5 m twice in an interval of 10 s. What is its time of flight?

A
28 s
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B
86 s
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C
104 s
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D
72 s
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Solution

## The correct option is C √104 sWe know that →s=→ut+12→at2 Note: in our sign convention upward is taken as positive⟹a=−g=−10m/s2 (as gravity acts downward)Now, 5=ut+12(−10)t2 5=ut−5t2 5t2−ut+5=0It is a quadratic equation in t having roots t1,t2and difference of root t1−t2=10s is givenWe know that difference of roots can be calculated ast1−t2=√u2−4×5×55=10⟹√u2−100=50⟹u=√2600m/sWe know that Time of flight T=2ug⟹T=2√260010⟹T=√104s

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