The height of the projection above the cannon is 10.78 m.
The given parameters;
- angle of projection = 45.0°
- initial velocity, v₀ = 25 m/s
- horizontal distance of the projectile, x = 50 m
The time of the motion of the projectile is calculated as;
[tex]X = v_0_x \times t\\\\X = (v_0\times cos(\theta)) \times t\\\\50 = (25 \times cos(45))t\\\\50 = 17.678t\\\\t = \frac{50}{17.678} \\\\t = 2.83 \ s[/tex]
The height above the cannon the net should be projected is calculated as;
[tex]h_y = v_0_yt - \frac{1}{2} gt^2\\\\h_y = (v_o \times sin(\theta))t\ - \ \frac{1}{2} gt^2\\\\h_y = (25\times sin(45)) \times 2.83 \ - \ (0.5 \times 9.8 \times 2.83^2)\\\\h_y = 50.02 - 39.24 \\\\h_y = 10.78 \ m[/tex]
Thus, the height of the projection above the cannon is 10.78 m.
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