Answer:
[tex]16.25^{\circ}[/tex]
Explanation:
R = Horizontal range of projectile = 75 m
v = Velocity of projectile = 37 m/s
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Horizontal range is given by
[tex]R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}[/tex]
The angle at which the arrow is to be released is [tex]16.25^{\circ}[/tex].
