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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<h3>a)</h3>
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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<h3>b)</h3>
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 <em>over</em> the branch.
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<em>Additional comment</em>
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).
