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Assume that a ball is kicked at an angle of `60^(@)` with the horizontal, so if the horizontal component of its velocity is `19.6 ms^(-1)`, determine its maximum height. |
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Answer» Given, `theta = 60^(@)` Horizontal component of velocity `= u cos 60^(@) = 19.6 ms^(-1)` `:. U = (19.6)/(cos 60^(@)) = (19.6)/(0.5) = 39.2 ms^(-1)` Therefore, maximum height, `H = (u^(2) sin^(2) 60^(@))/(2g) = ((39.2)^(2))/(2 xx 9.8) xx ((sqrt(3))/(2))^(2) = 58.8m` |
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