( Figure 1 ) A relief airplane is delivering a food package to a group of people

Question

(Figure 1) A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 482km/hour at an altitude of 725m . The positive x and y directions are defined in the figure. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80m/s2 .

A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 482km/hour at an altitude of 725m . The positive x and y directions are defined in the figure. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80m/s2 . After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 482 mph in the horizontal direction. If the package is to land right on the island, at what horizontal distance D from the plane to the island should the package be released? What is the speed of the package when it hits the ground?

Explanation / Answer

This is a kinematics problem (motion described by time, position, velocity, and acceleration), so start off by going to the correct equations in your text (probably in the same chapter as the problem?).

A) The key is to realize that in the vertical direction the object has zero initial velocity. The equation needed is:

yf = yi + vi*t + 1/2*a*t^2 where yf and yi are initial and final elevations, vi is initial velocity (zero in this case), and a is acceleration, in this case gravity. Make your point of zero altitude your origin, with the positive direction pointing up (this means yf = 0, and acceleration has to be taken as NEGATIVE when you plug it in because gravity acts downward). Plug in your numbers and solve for t.

B) The important thing to note here is that the acceleration in the horizontal direction is zero for projectile motion problems, and the initial velocity is the same as that of the plane. So again your equation is:

xf = xi + vi*t +1/2*a*t^2 where vi this time refers to the speed of the plane. Taking your horizontal origin to be at the point of release (xi = 0) and simplifying gives

xf = vi*t

You know t from part A, plug in and solve.

C) Now that you know everything, this is the easiest part. Acceleration is simply change in velocity over time. Your governing equation (in the vertical direction, which is what the question is asking) is:

vf = vi + a*t where vi =0 (vertical direction). Direction does not matter in this particular case, so take gravity to be positive.

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