A) 10 : 49
B) 49 : 10
C) 98 : 10
D) 10 : 98
Correct Answer: B
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 \[T=\frac{\sqrt{2}u}{g}\] Now, dividing R by \[{{T}^{2}}\], we get \[\frac{{{u}^{2}}}{g}\div \frac{2{{u}^{2}}}{{{g}^{2}}}=\frac{g}{2}=\frac{49}{10}\]
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