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A ball is kiched at an angle ` 30^(@)` with the verical. If the horizontal componet of its velocity is `20 ms^(-1)`, find the maximum hight and hrizontal range. Use `= 10 ms^(-2)`. |
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Answer» Here, theta =90^(@)-30^(@) =60^(@)` Horizontal velocity, ` u cos 60^(@) =20 ms^(-1)` or `u =(20)/( cos 60^(@)) =(20)/(1//2) =40 ms^(-1)` Maximum height , `H=(u^(2) sin^(2) theta ) /(2 g) =(40^(2) sin ^260^(@))/(2 xx 10)` `=(1600)/(2 g) =(40^(2) sin ^(2) 60^(@))/(10)` `=(17600)/(10) (sqrt 3//2) =238.6 m` |
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