question_answer

A projectile is thrown in the upward direction making an angle of\[60{}^\circ \]with the horizontal direction with a velocity of\[147\text{ }m{{s}^{-1}}\]. Then the time after which its inclination with the horizontal is\[45{}^\circ ,\]is

A)  15s                                       

B)  10.98s

C)  5.49s                                    

D)  2.745s

Correct Answer: C

Solution :

                 Initial velocity of projectile,\[u=147\text{ }m/s\] At the two points of the trajectory during projection, the horizontal component of the velocity is the same. \[u\cos {{60}^{o}}=v\cos {{45}^{o}}\] \[147\times \frac{1}{2}=v\times \frac{1}{\sqrt{2}}\] \[v=\frac{147}{\sqrt{2}}m/s\] Vertical component of \[u=u\sin {{60}^{o}}\]                                 \[=\frac{147\sqrt{3}}{2}m\] Vertical component of\[v=v\sin {{45}^{o}}\]                                 \[=\frac{147}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\]                                 \[=\frac{147}{2}m\] \[{{v}_{y}}={{u}_{y}}-gt\]                 \[\frac{147}{2}=\frac{147\sqrt{3}}{2}-9.8t\] \[9.8t=\frac{147}{2}(\sqrt{3}-1)\]                 \[t=\frac{147}{2\times 9.8}(\sqrt{3}-1)\]                 \[=5.49s\]