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A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the ball, with the same speed ? |
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Answer» Let u be the velocity of projection of the ball. The ball will cover maximum horizontaal distance when angle of projection with horizontal , `theta = 45^(@)`. Then `R_(max) = (u^(2))/(g) = 100 ` m If ball is projected vertically upwards `(theta =90^(@) ` from ground `)` then H attains maximum value, `H_(max) =(u^(2))/(2g) = (R_(max))/(2)` `therefore ` the height to which cricketer can throw the ball is `= (R_(max))/(2) = (100)/(2) = 50 `m |
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