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Two balls are dropped from different heights at different instants. Seconds ball is dropped 2 sec. after the first ball. If both balls reach the ground simultaneousl after 5 sec. of dropping the first ball. then the difference of initial heights of the two balls will be :- `(g=9.8m//s^(2))`A. 58.8mB. 78.4mC. 98.0mD. 117.6m |
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Answer» Correct Answer - B `h_(1)-h_(2)=(1)/(2)g(t_(1)^(2)-t_(2)^(2))=(1)/(2)xx9.8(5)^(2)-(3)^(2)` `=8xx9.5=78.4m` |
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