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 Time of rise \[{{t}_{1}}=\frac{u}{g+a}\]and height reached \[=\frac{{{u}^{2}}}{2(g+a)}\] Time of fall \[{{t}_{2}}\] is given by \[\frac{1}{2}(g-a)t_{2}^{2}=\frac{{{u}^{2}}}{2(g+a)}\] \[\Rightarrow {{t}_{2}}=\frac{u}{\sqrt{(g+a)(g-a)}}=\frac{u}{(g+a)}\sqrt{\frac{g+a}{g-a}}\] \[\therefore \]\[{{t}_{2}}>{{t}_{1}}\] because \[\frac{1}{g+a}<\frac{1}{g-a}\]
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