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A boy stands at 78.4 m away from a bulki...

A boy stands at 78.4 m away from a bulking and throws a ball which just enters a window at maximum height 39.2 m above the ground. Calculate the velocity of projection of the ball.

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Maximum height H = `(u^(2)sin^(2) theta )/( 2g) = 39.2m…….. `(i) Range = `(u^(2) sin 2 theta)/( g) = (2u^(2)sin theta cos theta)/(g) = 2 xx 78.4 ….. `(ii)
from equation (i) divided by equation (ii) `tan theta = 1rArr theta =45^(@)`
from equation (ii) range = `(u^(2) sin 90^(@))/(g) = 2xx 78.4 rArr u= sqrt( 2 xx 78.4 xx 9.8) = 39. m//s`
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