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 6 with the horizontal. The vertical velocity at 0 is zero. From       \[{{v}^{2}}={{u}^{2}}-2gh\] \[v={{v}_{y}}=0\]and \[u={{u}_{y}}=u\text{ }sin\,\theta \]                 \[0={{(u\sin \theta )}^{2}}-2gH\] \[\Rightarrow \]               \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Also, range = horizontal velocity\[\times \]time of flight                 \[R={{u}_{x}}\times T=(u\cos \theta )\times \frac{2u\sin 2\theta }{g}\]                 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] Given,   \[H=R\] \[\therefore \]  \[\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}\sin 2\theta }{g}\]                 \[\frac{{{\sin }^{2}}\theta }{\cos \theta }=4\] \[\Rightarrow \]               \[\tan \theta =4\] \[\therefore \]  \[\theta ={{\tan }^{-1}}(4),\theta \approx {{76}^{o}}\]