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A body is thrown horizontally with a velocity sqrt(2gh) from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is : |
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Answer» Solution :Here `x=vt` Now `h=1/2g t^(2)` or `t=sqrt((2h)/G)` THUS `x=sqrt(2gh)xxsqrt((2h)/g)=2h` |
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