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When a particle is projected at an angle...

When a particle is projected at an angle to the horizontal, it has range R and time of flight `t_(1)`. If the same projectile is projected with same speed at another angle to have the saem range, time of flight is `t_(2)`. Show that:
`t_(1)t_(2)=(2R//g)`

Text Solution

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The correct Answer is:
2
[2]: When particle is projected with same speed at angles `theta and 90-theta`, then its rnge is found tio be same for both cases.
`R=(u^(2) sin2theta)/(g)`
`t_(1)=(2u sin theta)/(g)`
`rArr t_(2)=(2u sin (90-theta))/(g)=(2 u cos theta)/(g)`
`t_(1)t_(2)=2(u^(2)(2sin theta cos theta))/(g^(2))=(2)/(g)(u^(2)sin 2theta)/(g)=(2R)/(g)`
On comparing with the given result we get n=2.
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