A) \[5\sqrt{\frac{g}{2}}\]
B) \[5\frac{g}{\sqrt{2}}\]
C) \[\frac{1}{3}\frac{g}{\sqrt{2}}\]
D) \[\frac{1}{5}\sqrt{\frac{g}{2}}\]
Correct Answer: A
Solution :
Given: Height of projectile H = 4m Horizontal range R = 12 m Height of projectile is \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ?(i) Pange of projectile is \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] ?(ii) Dividing equation (i) by (ii), also putting given value, we get \[=\frac{H}{R}=\frac{{{u}^{2}}\sin \theta }{2g}/\frac{{{u}^{2}}\sin 2\theta }{g}\] \[=\frac{{{u}^{2}}\sin \theta }{2g}\times \frac{g}{{{u}^{2}}2\sin \theta \cos \theta }\] \[\frac{4}{12}=\frac{\tan \theta }{4}\]or \[\tan \theta =\frac{4\times 4}{12}=\frac{4}{3}\] So, \[\sin \theta =\frac{4}{5}\] Now, putting \[H=4\,m\]and \[\sin \theta =\frac{4}{5}s\]in Eq. (i) we get \[H=\frac{{{u}^{2}}\times \frac{16}{5}}{2g}\]or \[4=\frac{{{u}^{2}}\times 16/25}{2g}\] \[{{u}^{2}}=4\times 2g\times \frac{25}{16}\] Hence, \[u=5\sqrt{\frac{g}{2}}m/s\]
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