Two persons simutaneously aim and fire their guns at a bird sitting on a tree. The first person fires his shot with a speed of `100m//s` at an angle of projection of `30^(@)`. The second person is ahead of the first by a distance of `50m` and fire his shot with a speed of `80m//s`. how must he aim his gun so that both the shots hit the bird simultaneously. Calculate the distance of the foot of the frre from two persons and the height of the tree. With what velocities and when do the two shorts hit the bird.

Explanation

The situation of the problem is shown in the fig.

For first short:
Let horizontal displacement `= x_(1)`
Vertical displacement = `h`
For second short:
Horizontal displacement `= x_(2)`
Vertical displacement `= h`
Now it is given `x_(1) = x_(2) +50`
For first shot, `h = (100sin 30) t-(1)/(2)"gt"^(2)`
For second shot `h = (80 sin theta)t - (1)/(2)"gt"^(2)`
or `sin theta = (100sin 30)/(80) = (50)/(80) = 0.625`
`:. theta = sin -10.625 = 38^(@)68'`
Now, `x_(1) = (100 cos 30^(@))t`
and `x_(2) = (80 cos 38^(@)68')t`
`:. (100 cos 30^(@))t =80 cos (38^(@)68')t +50`or
`t[100(sqrt(3)//2) -80 xx 0.7806] = 50`
or `t = (50)/(24.1526) = 2.070sec.`
Further `x_(1) = 100 cos 30^(@) xx 2.070 = 179.27m`
and `x_(2) = 179.27 - 50 = 129.27m`
or `h = (100 sin 30^(@)) (2.07)-((1)/(2)) xx 9.8 xx (2.07)2`
`=82.504m`
`KE = (1)/(2)MV_(cm)^(2) +(1)/(2)I_(cm)omega^(2)`