question_answer

The angle for which maximum height and horizontal range are same for a projectile is:

A) \[32{}^\circ\]                       

B) \[48{}^\circ\]

C) \[76{}^\circ\]                              

D) \[84{}^\circ\]

Correct Answer: C

Solution :

Key Idea: Vertical component of velocity at the highest point is zero. Let body is projected with an initial velocity u making an angle \[\theta \] with the horizontal. The vertical velocity at O is zero. From       \[{{v}^{2}}={{u}^{2}}-2gh\] \[v={{v}_{y}}=0\]and \[u={{u}_{y}}=u\sin \theta \] \[0={{(u\,sin\theta )}^{2}}-2\,gH\] \[\Rightarrow \]\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Also, range = horizontal velocity \[\times \]time of flight \[R={{u}_{x}}\times T=(u\,cos\,\theta )\times \frac{2u\,\sin \theta }{g}\] \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] Given,         \[H=R\] \[\therefore \]       \[\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[\frac{{{\sin }^{2}}\theta }{2}=2\sin \theta \cos \theta \] \[\frac{\sin \theta }{\cos \theta }=4\] \[\Rightarrow \]      \[\tan \theta =4\] \[\therefore \]    \[\theta ={{\tan }^{-1}}(4)\theta \approx {{76}^{o}}\]