Answer:19.09 m
Explanation:
Given
[tex]\theta =48.3^{\circ}[/tex]
initial Velocity(u)=32.5 m/s
Distance given(x)=32.8
We know equation of trajectory of Projectile is given by
[tex]y=x\tan \theta -\frac{gx^2}{2u^2cos^2\theta }[/tex]
[tex]y=32.8\times \tan 42.8-\frac{9.8(32.8)^2}{2(32.5)^2\times (cos48.3)^2}[/tex]
y=30.37-11.278=19.09 m
thus Net must be kept at a vertical distance of 19.09 m above canon