question_answer

A projectile is thrown with initial velocity \[\text{(a}\overset{\hat{\ }}{\mathop{\text{i}}}\,+\text{b}\overset{\hat{\ }}{\mathop{\text{j}}}\,\text{)m}/\text{s}.\]If range of projection is twice the maximum height reached by it, then

A)  \[b=\frac{a}{2}\]                            

B)  \[b=a\]

C)  \[b=2a\]                             

D)  \[b=4a\]

Correct Answer: C

Solution :

                 The horizontal range (R) is given by                                 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\]                      ?. (i) and        \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]                       ... (ii) Given,   \[u=a\,\hat{i}+b\,\hat{j}={{u}_{x}}\,\hat{i}+{{u}_{y}}\,\hat{j}\] \[\therefore \]  \[{{u}_{x}}=u\cos \theta =a\]                     ... (iii)                 \[{{u}_{y}}=u\sin \theta =b\]                      ... (iv) \[\therefore \]  \[R=\frac{{{u}^{2}}(2\,\sin \theta )\,\,(\cos \theta )}{g}=\frac{2\,ab}{g}\]                 \[H=\frac{{{(u\sin \theta )}^{2}}}{2g}=\frac{{{b}^{2}}}{2\,g}\] Given,   \[R=2\,H\]                 \[\frac{2\,ab}{g}=2\,\frac{{{b}^{2}}}{2\,g}\] \[\Rightarrow \]               \[b=2\,a\]