question_answer

Neglecting the air resistance, the time of flight of a projectile is determined by :

A) \[{{U}_{vertical}}\]                                         

B) \[{{U}_{horizontal}}\]

C) \[U=\mathsf{U}_{vertical}^{2}+\mathsf{U}_{horizontal}^{2}\] 

D)  \[U={{(\mathsf{U}_{vertical}^{2}+\mathsf{U}_{horizontal}^{2})}^{{}^{1}/{}_{2}}}\] 

Correct Answer: A

Solution :

                Let a body be projected at an initial velocity\[u\] in a direction making an angle\[\theta \]with the horizontal and let it take time t to reach the highest point P of its path. The vertical velocity of the body at P is zero. From equation of motion. \[v=u-gt\] Putting, \[v={{v}_{y}}=0\]and \[u={{u}_{y}}=usin\theta \] \[0=u\text{ }\sin \theta -gt\] \[\Rightarrow \]               \[t=\frac{u\sin \theta }{g}\] Hence, time of flight (T) is 2t. \[\therefore \]  \[T=2t=\frac{2u\sin \theta }{g}\] \[=\frac{2}{8}\times {{U}_{vertical}}\]