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Feliz [49]
3 years ago
13

an archer stands 40.0m from the target. if the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull's

eye must she aim to compensate for gravity pulling her arrow downward? ...?
Physics
1 answer:
pshichka [43]3 years ago
3 0
Ok, first we need to have in mind that to findthe distance we need to find also the time then we use the kinematic equations in the x-direction 

<span>delta x = Vox*t + 1/2*a*t^2 </span>
<span>since there is no horizontal acceleration, a = 0 m/s/s </span>
<span>delta x = Vox*t </span>
<span>40.0 m = 90 m/s * t </span>
<span>t = 4/9 s </span>

<span>Now using this time you can calculate the vertical displacement due to gravity. </span>
<span>delta y = Voy*t + 1/2*g*t^2 </span>
<span>since there is no initial vertical velocity, Voy = 0m/s </span>
<span>delta y = 1/2*g*t^2 </span>
<span>delta y = 1/2*(-9.81 m/s/s)*(4/9 s)^2 </span>
<span>solve for delta y and you will have the answer</span>
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Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

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