A) \[10\sqrt{3}\,m{{s}^{-1}}\]
B) \[20\sqrt{3}\,m{{s}^{-1}}\]
C) \[10\sqrt{5}\,m{{s}^{-1}}\]
D) \[20\sqrt{2}\,m{{s}^{-1}}\]
Correct Answer: C
Solution :
Suppose the angle made by the instantaneous velocity with the horizontal be\[\alpha \]. Then \[\tan \alpha =\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{u\sin \theta -gt}{u\cos \theta }\] Given: \[\alpha =45{}^\circ ,\] when \[t=1\text{ }s\] \[\alpha =0{}^\circ ,\]when \[t=2\text{ }s\] This gives \[u\text{ }cos\theta =u\text{ }sin\theta -g\] and \[u\text{ }sin\theta -2g=0\] Solving we have, \[u\text{ }sin\theta =2g\] \[u\,cos\,\theta =g\] \[\therefore \] \[u=\sqrt{5}g=10\sqrt{5}m/s\]
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