Question

An archer shoots an arrow at an 85.0 m distant target; the bulls eye of the target is at same height as the release height of the arrow. at what angle in degrees mist the arrow be released to hit the bulls eye if its initial speed is 42.0 m/s?
there is a large tree halfway between the archer and the target with an overhanging branch 5.07 m above the release height of the arrow. will the arrow go over or under the branch?
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Answer 1

Answer:

  a) 14.1°

  b) over

Explanation:

The usual model of ballistic motion assumes that the only force on the flying object is that due to gravity. When an object is launched with initial velocity v0 at some angle θ with respect to the horizontal, the distance it travels is ...

  d = (v0)²sin(2θ)/g

Using this relation, we can find the launch angle to make the object travel a given distance:

  θ = 1/2arcsin(dg/v0²) . . . . where g is the acceleration due to gravity

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a)

For the arrow to hit a target 85 m away at the same height it was launched with speed 42.0 m/s, the launch angle must be ...

  θ = 1/2arcsin(dg/v0²) = 1/2(arcsin(85·9.8/42²)) ≈ 14.0893°

The arrow must be released at an angle of about 14.1°.

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b)

The flight time to the tree at a distance of 42.5 m will be that distance divided by the horizontal speed:

  t = 42.5/(42cos(14.0893°)) ≈ 1.0433 . . . . seconds

The height at that time is ...

  h(t) = -4.9t² +42sin(14.0893°)t ≈ 5.33 . . . meters

The arrow will go over the branch.

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Additional comment

Since gravity provides the only force on the arrow, its horizontal speed is constant at vh = v0·cos(θ), when the arrow is launched with speed v0 at angle θ above the horizontal. Its vertical speed will be reduced by the acceleration of gravity, so will be vv = v0·sin(θ) -gt. The height is the integral of the vertical speed, so is ...

  h(t) = (1/2)gt² +v0·sin(θ)t

The height will be 0 at t=0 and at t=2v0sin(θ)/g, so the horizontal distance traveled will be ...

  d = vh·t

  = (v0·cos(θ))(2v0·sin(θ)/g) = (v0²/g)(2·sin(θ)cos(θ))

  = v0²sin(2θ)/g

Note that this is all simplified by the fact that the target and launch point are at the same level (h=0).