A body stans at `78.4 m` from a building and throws a ball which just enters a window `39.2 m` above the ground. Calculate the velocity of projection of the ball. Fig. 2 (d) . 22.
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Explanation

Maximum height `=(u^(2)sin^(2)theta)/(2g)=39.2m`...(i) Range `=(u^(2)sin2 theta)/(g)=(2u^(2)sin thetacostheta)/(g)=2xx78.4` ....(ii)
from equation (i) divided by equation (ii) `tan theta=1rArrtheta=45^(@)`
from equation (ii) range `=(u^(2)sin90^(@))/(g)=2xx78.4rArru=sqrt(2xx78.4xx9.8)=39.2 m//s`.