An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight? 1)28 0.5 2)86 0.5 3)104 0.5 4)72 0.5

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bharat bajaj · Answer

s = ut + 1/2. a.t^2 ..... 5 = ut - 5t^2 ..... 5t^2 - ut + 5 = 0 ... The roots of this equation gives us the time at which particle crosses the height 5 m....If we calculate the difference of roots, it tells us the interval which we know is 10 s... Say roots are a and b....a-b = ( (a+b)^2 - 4ab ) ^0.5....a - b = ( u^2/25 - 4)^0.5....100 = u^2/25 - 4.. u = 5.(104)^0.5...Total time of flight = 2u/a = (104)^0.5..THanks Bharat BajajIIT Delhi askiitians faculty

Arun Kumar · Answer

Hello Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty

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An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight? 1)28 0.5 2)86 0.5 3)104 0.5 4)72 0.5

An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight?
1)28^0.5
2)86^0.5
3)104^0.5
4)72^0.5

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An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight? 1)28 0.5 2)86 0.5 3)104 0.5 4)72 0.5

An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight?1)280.52)860.53)1040.54)720.5

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An object is thrown vertically up from the ground passes the height 5 m twice in an interval of 10 secs. what is the time of flight?
1)28^0.5
2)86^0.5
3)104^0.5
4)72^0.5