question_answer

A projectile is thrown in the upward direction making an angle of \[2mA\] with the horizontal direction with a velocity of 147 m/s. Then, the time after which its inclination with the horizontal is \[3.6mA\], is

A)  \[1.8mA\]                                          

B)  \[62V,\,\,2\Omega \]

C)  \[63V,\,\,1\Omega \]                   

D)  \[61V,\,\,1\Omega \]

Correct Answer: C

Solution :

                Initial velocity of projectile, \[\text{4 }km/h\] At the two points of the trajectory during projection, the horizontal component of the velocity is the same.                 \[\text{5 }km/h\]                 \[{{10}^{-6}}{{m}^{2}}\]                 \[{{10}^{29}}\] Vertical component of \[250\times {{10}^{-3}}m/s\]                                                 \[25\times {{10}^{-3}}m/s\] Vertical component of \[2.50\times {{10}^{-3}}m/s\]                 \[1.25\times {{10}^{3}}m/s\] \[\frac{hv}{4}\] \[\frac{hv}{3}\] From the equation                 \[\frac{hv}{2}\]                 \[\frac{2hv}{3}\]                 \[30\mu A\]                 \[90\mu A\]                 \[=5.49s\]