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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)280.52)860.53)1040.54)720.5`
5 years ago

bharat bajaj
IIT Delhi
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..THanksBharat BajajIIT Delhiaskiitians faculty
```
5 years ago
Arun Kumar
IIT Delhi
256 Points
```							Hello$\\ \\T_{flight}={2 v \over g} \\=>v={gT_{flight} \over 2} \\=>v=50m/s  at height 5m \\v^2=50^{2}+2gs=2500+100=2600 \\=>v=\sqrt{2600} \\=>T={2 v \over g}=2\sqrt{26} \\$Thanks & RegardsArun KumarBtech, IIT DelhiAskiitians Faculty
```
5 years ago
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