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Grade: 11
        
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)280.5
2)860.5
3)1040.5
4)720.5
5 years ago

Answers : (2)

bharat bajaj
IIT Delhi
askIITians Faculty
122 Points
							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
5 years ago
Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							
Hello
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
5 years ago
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