Question

For a given velocity, a projectile has the same range R for two angles of projection. If t₁ and t₂ are the time of flight in the two cases, then t₁t₂ is equal to

Moderate

Solution

R $= \frac{u^{2} \sin 2 θ}{g} at angles θ and 90 ° - θ$
Now, $t_{1} = \frac{2 u \sin θ}{g} and t_{2} = \frac{2 u \sin (90 ° - θ)}{g} = \frac{2 u \cos θ}{g}$
$∴ t_{1} t_{2} = \frac{2}{g} (\frac{u^{2} \sin 2 θ}{g}) = \frac{2 R}{g}$

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