Answer:30.34 m
Explanation:
Given
launch angle[tex]=51.7^{\circ}[/tex]
Initial velocity u=31.9 m/s
Position of net = 39.1 m
We know equation of trajectory of a projectile is
[tex]y=x\tan \theta -\frac{gx^2}{2u^2(\cos \theta )^2}[/tex]
[tex]y=39.1\times \tan 51.7-\frac{9.8\times 39.1^2}{2\times 31.9^2\times (cos51.7)^2}[/tex]
[tex]y=49.509-\frac{9.8\times 1.502}{2\times 0.384}[/tex]
[tex]y=49.509-\frac{14.719}{0.768}[/tex]
[tex]y=49.509-19.165=30.343 m[/tex]
Therefore Net should be placed 30.34 m above ground
