A) equal to the time of fall
B) less than the time of fall
C) greater than the time of fall
D) twice the time of fall
Correct Answer: B
Solution :
Let the initial velocity of ball be u \[\therefore \]Time of rise \[{{t}_{1}}=\frac{u}{g+a}\] and height reached \[\text{= }\frac{{{u}^{2}}}{2\left( g+a \right)}\] Time of fall \[{{\text{t}}_{2}}\]is given by \[\frac{1}{2}\left( g-a \right)t_{2}^{2}=\frac{{{u}^{2}}}{2\left( g+a \right)}\] \[{{t}_{2}}=\frac{u}{\sqrt{\left( g+a \right)\left( g-a \right)}}=\frac{u}{\left( g+a \right)}\sqrt{\frac{g-a}{g+a}}\] \[\therefore \text{ }{{\text{t}}_{2}}>{{\text{t}}_{\text{1}}}\text{ because }\frac{1}{g+a}<\frac{1}{g-a}\]
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