Answer:
Explanation:
Given
velocity of launch [tex]u=50\ m/s[/tex]
Target is [tex]R=65\m\ away[/tex]
Suppose [tex]\theta [/tex] is the launch angle
We know Range of Projectile is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
[tex]65=\frac{50^2\times \sin 2\theta }{9.8}[/tex]
[tex]\sin 2\theta =0.254[/tex]
because [tex]\sin \theta =\sin (180-\theta )[/tex]
so either [tex]2\theta =14.71[/tex]
[tex]\theta =7.35^{\circ}[/tex]
or [tex]180-2\theta =14.71[/tex]
[tex]\theta =82.64^{\circ}[/tex]
