question_answer

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

A)  \[{{U}_{vertical}}\]

B)  \[{{U}_{horizontal}}\]

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

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

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 \[u={{u}_{y}}=u\,\sin \theta \] \[0=u\,\sin \theta -gt\] \[\Rightarrow \] \[t=\frac{u\,\,\sin \,\theta }{g}\] Hence, time of flight (T) is 2t. \[\therefore \] \[T=2t=\frac{2u\,\sin \theta }{g}\] \[=\frac{2}{g}\times {{U}_{vertical}}\]