A) \[15\,s\]
B) \[10.98\,s\]
C) \[5.49\,s\]
D) \[2.745\,s\]
Correct Answer: C
Solution :
\[u=147m/s,\] \[\theta =60{}^\circ \] Vertically, \[v\sin \phi =u\sin \theta -gt\] \[\Rightarrow \,\,9.8t=u\sin 60{}^\circ -v\sin 45{}^\circ \] \[=\frac{u\sqrt{3}}{2}-\frac{v}{\sqrt{2}}\] ?.(i) Horizontally, \[u\cos \theta =v\cos \phi \] \[\Rightarrow \,\,u\cos 60{}^\circ =v\cos 45{}^\circ \] or \[v=\frac{u}{\sqrt{2}}\] ?..(ii) \[\therefore \,\,t=\frac{1}{2(9.8)}(\sqrt{3}-1)147=5.49\,\sec \]You need to login to perform this action.
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