A) \[15(\sqrt{3}-1)\,s\]
B) \[15(\sqrt{3}+1)\,s\]
C) \[7.5(\sqrt{3}-1)\,s\]
D) \[7.5(\sqrt{3}+1)\,s\]
Correct Answer: C
Solution :
[c] At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then \[u\,\,\cos \,\,60{}^\circ =v\,\,\cos \,\,45{}^\circ \] \[\Rightarrow \]\[150\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\] or \[v=\frac{150}{\sqrt{2}}\,\,m/s\] Initially: \[{{u}_{y}}=u\,\sin \,60{}^\circ =\frac{150\sqrt{3}}{2}\,\,m/s\] Finally: \[{{v}_{y}}=v\,\sin \,45{}^\circ =\frac{150}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{150}{2}\,\,m/s\] But \[{{v}_{y}}={{u}_{y}}+{{a}_{y}}t\] or \[\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t\] \[10t=\frac{150}{2}\left( \sqrt{3}-1 \right)\] or \[t=7.5\left( \sqrt{3}-1 \right)s\]
You need to login to perform this action.
You will be redirected in
3 sec
