A) \[\frac{{{p}^{2}}}{4q}\]
B) \[\frac{{{p}^{2}}}{2q}\]
C) \[\sqrt{\frac{2p}{qg}}\]
D) \[p\sqrt{\frac{2}{qg}}\]
Correct Answer: D
Solution :
[d]: Comparing the equation, \[y=px-q{{x}^{2}}\] with the equation of projectile motion\[y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }\]we get, \[\tan \theta =p\]and \[\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }=q\]or\[u\cos \theta =\sqrt{\frac{g}{2q}}\] The time of flight of the projectile is \[T=\frac{2u\sin \theta }{g}=\frac{2u\sin \theta \cos \theta }{g\cos \theta }=\frac{2u\cos \theta \tan \theta }{g\theta }\] \[=\frac{2}{g}(ucos\theta )(tan\theta )=\frac{2}{g}\left( \sqrt{\frac{g}{2q}} \right)(p)=p\sqrt{\frac{2}{qg}}\]
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