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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 s
We know that s=ut+12at2 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=ut5t2
5t2ut+5=0
It is a quadratic equation in t having roots t1,t2
and difference of root t1t2=10s is given
We know that difference of roots can be calculated as
t1t2=u24×5×55=10
u2100=50
u=2600m/s

We know that Time of flight T=2ug
T=2260010
T=104s



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