1. A jet is flying with an air speed of 500 m/s. A parcel is dropped out of the plane. Assume that it takes 40 seconds for the parcel to hit the ground.. Assume no air resistance.
a. How high is the jet?
b. How far forward horizontally from the spot where it leaves the jet, does the parcel move during those 40 seconds?
c. Where is the parcel relative to the jet?
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To find the answers to these questions, we can use equations of motion.
a. To determine the height of the jet, we need to find the time it takes for the parcel to fall. We know that the time is 40 seconds. Since the parcel is dropped, its initial velocity is 0 m/s. The acceleration due to gravity can be assumed to be approximately 9.8 m/s^2. We can use the equation of motion, s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we have:
s = 0 * 40 + (1/2) * 9.8 * (40)^2
s = 0 + 1/2 * 9.8 * 1600
s = 1/2 * 9.8 * 1600
s = 7840 m
Therefore, the height of the jet is 7840 meters.
b. To find how far forward the parcel moves horizontally during those 40 seconds, we need the horizontal component of the jet's velocity. However, the question only gives the airspeed, which is the magnitude of the velocity vector. We don't have the angle. Hence, we cannot determine the horizontal distance traveled by the parcel.
c. Since the parcel is dropped out of the plane, it falls freely under the influence of gravity. Therefore, relative to the jet, the parcel is simply falling downward.