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Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is t_(1) = 2s and for the second t_(2) = 1s. At what height was the first body situated when the other began to fall? (g=10 m//s^(2)) |
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Answer» Solution :The second body takes 1 second to reach ground. So, we need to find the DISTANCE TRAVELED by the first body in its first second and in two seconds. The distance covered by first body in 2s, `h_(1)=1//2 GT^(2) =1//2 xx10xx2^(2)=20 m`. The distance covered in `1s, h_(2)=5 m`. The height of the first body when the other begin to fall H =20-5 =15 m. |
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