question_answer

A ball is thrown up, it reaches a maximum height and then comes down. If \[{{t}_{1}}\]and \[({{t}_{2}}>{{t}_{1}})\]are the times that the ball takes to be at a particular height then the time taken by the ball to reach the highest point is

A)  \[{{t}_{1}}+{{t}_{2}}\]

B)  \[{{t}_{2}}-{{t}_{1}}\]

C)  \[\frac{{{t}_{2}}-{{t}_{1}}}{2}\]

D)  \[\frac{{{t}_{1}}+{{t}_{2}}}{2}\]

Correct Answer: D

Solution :

: Time for journey\[AB={{t}_{1}}\]             ...(i) Time for journey\[ABCB={{t}_{2}}\] \[\therefore \] Time for \[(AB+BC+CB)={{t}_{2}}\]        ...(ii) Add up (i) and (ii) Time for \[(AB)+(AB+BC+CB)={{t}_{1}}+{{t}_{2}}\] or Time for \[2AB+2BC={{t}_{1}}+{{t}_{2}}\] or Time for\[(AB+AC)=\frac{({{t}_{1}}+{{t}_{2}})}{2}\] \[\therefore \]Time for reaching highest point\[=\frac{({{t}_{1}}+{{t}_{2}})}{2}\]