Question

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

• A. 10 : 49
• B. 49 : 10
• C. 98 : 10
• D. 10 : 98

Answer 1

Answer: (B)

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 $\frac{u^2}{g}+\frac{2u^2}{g^2}=\frac{g}{2}=\frac{49}{10}$