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InterviewSolution
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A ball is thrown up with a speed of 15 m/s . How high wiil it go before it begins to fall ? `(g=9.8 m//s^(2)` |
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Answer» Please note that here the ball is going up against the gravity ,so the value of g is to be taken as negative . Here Initial speed of ball u=15 /s Final speed of ball ,v=0 (The ball stops ) Acceleration due to gravity `g= -9.8 m//s^(2)` (Retardation ) And Height h=? (to be calculated ) Now ,Putting al these values in the formula `v^(2)=u^(2)+2gh ` we get : `(0)^(2)=(15)^(2)+2xx(-9.8) xx h` 0=225-19.6 h 19.6 h =225 `h=225/19.6` h=11.4 m Thus the ball will go the a maximum height of 11.4 meters before of begins to fall. |
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