A) 5.7 m/s
B) 7.7 m/s
C) 36.1 m/s
D) 9.7 m/s
Correct Answer: B
Solution :
\[\tan \theta =\frac{{{v}^{2}}}{rg}\] \[\sin \theta =\frac{{{v}^{2}}}{rg}\] \[(\sin \theta \approx \tan \theta ,for\,small\,\theta )\] \[\therefore \] \[\frac{1.2}{8}=\frac{{{v}^{2}}}{40\times 9.8}\] \[{{v}^{2}}=1.2\times 49\] \[v=\sqrt{1.2\times 49}\] \[=\sqrt{58.8}=7.7\,m/s\]You need to login to perform this action.
You will be redirected in
3 sec