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AVprozaik [17]
3 years ago
9

An arrow is launched from a bow with an initial horizontal velocity of 40

Physics
1 answer:
Ivan3 years ago
4 0

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

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   $\vec{F}_B = $ force applied by Barry

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The resultant,

$\vec{F}_R= \vec{F}_A+\vec{F}_B $

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e). The direction of the force exerted by Eric is exactly opposite to the direction of the resultant force.

The direction of the resultant force is :

$\theta = \tan ^{-1}\left(\frac{F_y}{F_x}\right)$

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The direction of the force E is  45° west or  45° south west.

   

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