contestada

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the arrow. If d = 75 m and v = 37 m/s. At what angle must the arrow be released to hit the bull's-eye if its initial speed is 37 m/s?

Respuesta :

boffeemadrid

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].

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