Answer :
(a) The distance along the ground, measured from a point directly beneath the point of release to where the package hits the earth, is approximately 817.5 meters.
(b) The angle of the velocity vector of the package just before impact, relative to the ground, is approximately 40.9 degrees.
To calculate the distance along the ground (horizontal distance) covered by the package, we need to find the time it takes for the package to hit the ground after being released. Since the vertical motion is affected by gravity, we can use the kinematic equation:
[tex]Δy = Vyi * t + (1/2) * a * t^2[/tex]
where Δy is the vertical displacement, Vyi is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2). Given that the initial vertical velocity is determined by the upward velocity component of the airplane, Vyi = V * sin(θ), where V is the airplane's speed (97.6 m/s) and θ is the angle of climb (37.5 degrees).
Plugging in the values, we have:
[tex]654 m = (97.6 m/s * sin(37.5°)) * t + (1/2) * (-9.8 m/s^2) * t^2[/tex]
Simplifying the equation and solving for t, we find two possible solutions: t = 0 s (at the release point) and t ≈ 14.35 s. Since we're interested in the time it takes for the package to hit the ground, we consider the positive time value.
Next, to find the horizontal distance traveled by the package, we use the equation:
[tex]Δx = Vxi * t[/tex]
where Δx is the horizontal displacement and Vxi is the initial horizontal velocity. The initial horizontal velocity is given by Vxi = V * cos(θ).
Plugging in the values, we have:
Δx = (97.6 m/s * cos(37.5°)) * 14.35 s ≈ 817.5 m
Therefore, the distance along the ground, measured from a point directly beneath the point of release to where the package hits the earth, is approximately 817.5 meters.
To determine the angle of the velocity vector just before impact, we use trigonometry. The horizontal component of the velocity remains constant, Vxf = Vxi = V * cos(θ). The vertical component of the velocity can be determined using Vyf = Vyi + a * t, where Vyi is the initial vertical velocity and a is the acceleration due to gravity. Plugging in the values, we have Vyi = V * sin(θ) and Vyf = V * sin(θ) +[tex](-9.8 m/s^2)[/tex] * 14.35 s.
To find the angle, we use the tangent function:
θf = tan^(-1)(Vyf / Vxf)
Plugging in the values, we have [tex]θf = tan^(-1)((V * sin(37.5°) + (-9.8 m/s^2)[/tex] * 14.35 s) / (V * cos(37.5°))). Evaluating this expression gives us θf ≈ 40.9 degrees.
Therefore, the angle of the velocity vector of the package just before impact, relative to the ground, is approximately 40.9 degrees.
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