A) 10 m/s, towards east
B) 10 m/s, towards north
C) 10 m/s, towards south
D) 10 m/s, towards north-east
Correct Answer: A
Solution :
\[\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}\,t\]\ \[v=\sqrt{{{u}^{2}}+{{a}^{2}}{{t}^{2}}+2u\,at\cos \theta }\] \[v=\sqrt{200+100+2\times 10\sqrt{2}\times 10\times \cos 135}\]\[=10\,m/s\] \[\tan \alpha =\frac{at\sin \theta }{u+at\cos \theta }\frac{10\sin 135}{10\sqrt{2}+10\cos 135}=1\]\\[\alpha =45{}^\circ \] i.e. resultant velocity is 10 m/s towards East.You need to login to perform this action.
You will be redirected in
3 sec