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A person can throw a ball to a maximum horizontal distance of ` 90.m` Calculat the maximum vertcal heitht to which he can through the ball. Fiven ` g= 10 ms^(-2)`. |
Answer» Max. Horizontal distace . ` R_(max) = u^2/g = 90` or ` u^2 = 90 g= 90 xx 10 = 900 so `u=30 m//s` ` Max. vertical height, ` H = u^2 /(2g) = (900)/(2 xx 10) = 45 m` . |
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