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A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity `20 ms^-1`. The second stone will overtake the first after travelling a distance of `(g=10 ms^-2)`A. `13 m`B. `15 m`C. `11.25 m`D. `19.5 m` |
Answer» Correct Answer - C `1/2g t^2=20(t-1)+1/2g(t-1)^2` Solving this equation we get, `:.t=1.5s` Now, `d=20(t-1)+1/2g(t-1)^2` `=11.25 cm` |
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