question_answer

A ball is projected up at an angle 9 with horizontal from the top of a tower with speed V. It hits the ground at point A after time \[{{t}_{A}}\] with speed \[{{v}_{A}}\]. Now, this ball is projected at ' same angle and speed from the base of the tower (located at point P) and it hits ground at point B after time \[{{t}_{B}}\] with speed \[{{v}_{B}}\]. Then

A)  \[PA=PB\]

B)  \[{{t}_{A}}<{{t}_{B}}\]

C)  \[{{v}_{A}}<{{v}_{B}}\]

D)  Ball A hits the ground at an angle \[(-\theta )\] with horizontal

Correct Answer: C

Solution :

Consider, the two projectiles projected from \[O\]and P as shown in the figure. Range of both the particles \[R=\frac{{{v}^{2}}\,\,\sin \,2\theta }{g}\] \[\Rightarrow \] \[OA'=PB=R\] Clearly, velocity at the points A' and B will be same. Velocity of the ball following the trajectory OC? A?. Will increase after A' due to accelerated motion under gravity. Therefore, we can say that \[{{v}_{A}}>{{v}_{B}}\]