A) 9.8 m/s
B) 10 m/s
C) 5.8 m/s
D) 17.3 m/s
Correct Answer: D
Solution :
Let the velocity along x and y axes be \[{{v}_{x}}\] and \[{{v}_{y}}\] respectively. \[\therefore {{v}_{x}}=\frac{dx}{dt}and\,{{v}_{y}}=\frac{dy}{dy}\] From figure, \[\tan \alpha =\frac{y}{x}\] \[\Rightarrow y=x\tan \alpha \] Differentiating Eq. (i) w.r.t. \[t\], we get \[\frac{dy}{dt}=\frac{dx}{dt}\tan \alpha \] \[\Rightarrow \] \[{{v}_{y}}={{v}_{x}}\tan \alpha \] Here, \[{{v}_{x}}=10\,m/s\,,\,\alpha ={{60}^{o}}\] \[\therefore {{v}_{y}}=10\,\tan {{60}^{o}}=10\sqrt{3}=17.3\,m/s\]You need to login to perform this action.
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