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Q. Ratio between maximum range and square of time of flight in projectile motion is

VMMC MedicalVMMC Medical 2009

Solution:

The range of the projectile is maximum for the angle of projection $ \theta ={{45}^{o}} $ The maximum range of the projectile is $ {{R}_{\max }}=\frac{{{u}^{2}}}{g} $ When the range of the projectile is maximum the time of flight $ T=\frac{\sqrt{2u}}{g} $ Now, dividing R by $ {{T}^{2}}, $ we get $ \frac{{{u}^{2}}}{g}+\frac{2{{u}^{2}}}{{{g}^{2}}}=\frac{g}{2}=\frac{49}{10} $