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A ball is thrown vertically upwards from the ground and a student gazing out of the window sees it moving upward past him at `10ms^-1.` The window is at 15 m above the ground level. The velocity of ball 3 s after it was projected from the ground is [Take `g= 10 ms^-2`]A. `10 m//s,` upB. `20 m//s,` upC. `20 ms^-1,`downD. `10 ms^-1,` down |
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Answer» Correct Answer - A Total height=`15+(u^2)/(2g)` `=15+((10)^2)/(2xx10)` or `h=20m` initial velocity `u=sqrt2gh` `sqrt(2xx10xx20)` `=20 m//s` Now applying `v=u+at,` we have `v=(+20)+(-10)(3)` `:.` Velocity is `10m//s,` downwards. |
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