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An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. At what angle (in degrees) must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s?

Respuesta :

CombatGaming505

Answer:

Peanuts

Explanation:

Because peanuts

joseaaronlara

Answer:

The answer to your question is angle = 18.46°

Explanation:

Data

d = 75 m

v₀ = 35 m/s

α = ?

Formula

                     [tex]d = \frac{vo^{2}sin2\alpha}{g}[/tex]

solve for sin2α

                  sin 2α = [tex]\frac{dg}{vo^{2}}[/tex]

Substitution

                 sin 2α = [tex]\frac{75(9.81)}{35^{2}}[/tex]

Simplify

                sin 2α = [tex]\frac{735.75}{1225}[/tex]

Divide

                sin 2α = 0.600

Get sin⁻¹

                      2α = 36.9°

Divide by 2

                        α = 18.5°                

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