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

Lrt boy standing at (A) throw a ball with a velocity (u) at an angle `theta` with the building and the ball just enters window (W),
As the boy is at `78.4 m` from the building and the ball just enters the window `39.2 m` above the ground therefore
Max , height, `u^(2) sin^(2) theta)/(2 g) =39.2 m` ....(i)
and horizontal range,
`(u^(2) sin 2 theta)/g =2 xx 78.4 m` ...(ii)
Dividing (i) by (ii), we get
`(u^(2) sin^(2) theta)/ 2 g) g/(u^(2) 2 sin theta cos theta) =(39.2)/( 2 xx 78.4)`
or ` (sin ^(2) theta)/(2 xx 2 sin theta cos theta) =1/4`
or ` tan theta =1 or theta =45^(@)`
Substituting in (ii), we get
`(u^(2) sin 90^(@))/(9.8) =2 xx 78 .4`
or `u=zsqrt (2 xx 78.4 xx 9.8 ) =3 9 .2 ms^(-1)`.