A) \[\sqrt{2.5}m\]
B) \[1\,\,m\]
C) \[3\,\,m\]
D) \[1.5\,\,m\]
Correct Answer: A
Solution :
From equation of motion \[h=ut+\frac{1}{2}g{{t}^{2}}\] where\[u\] is initial velocity and \[t\] time. Since,\[u=0\] \[\therefore \] \[t=\sqrt{\frac{2h}{g}}=\frac{\sqrt{2\times 4.9}}{9.8}=1\,\,s\] The horizontal range of the drop\[=x\], then \[x=\left( \frac{{{v}_{t}}}{o} \right)t\] Also, \[\omega =\frac{\Delta \theta }{\Delta t}=\frac{21\times 2\pi }{44}=3\,\,rad/s\] Tangential speed\[{{v}_{t}}=r\omega =0.5\times 3\times 1.5\,\,m/s\] \[x=1.5\times 1=1.5m\] Locus of drop\[=\sqrt{{{x}^{2}}+{{r}^{2}}}=\sqrt{{{(1.5)}^{2}}+{{(0.5)}^{2}}}\] \[=\sqrt{2.5}m\]
You need to login to perform this action.
You will be redirected in
3 sec
