A) 12s
B) 6s
C) 2s
D) 1s
Correct Answer: C
Solution :
\[U=10m/s\] Time taken to the ball thrown downwards to reach the surface is \[\frac{u}{g}={{T}_{1}}\] And time taken to the ball thrown up to reach the max. height and come to point of projection is\[\frac{2u}{g}\]and to reach the ground it takes u / g . Total time\[\frac{u}{g}+\frac{2u}{g}={{T}_{2}}.\] \[{{T}_{2}}-{{T}_{1}}=\frac{u}{g}+\frac{2u}{g}-\frac{u}{g}=\frac{2\times 10}{10}=2s\]You need to login to perform this action.
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