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 solve these questions, we can apply basic physics principles and equations. Let's break it down step by step:
a. To find the height of the jet, we need to determine how far the parcel falls during the given time. We can use the equation of motion for free-falling objects:
h = (1/2) * g * t^2
Where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken (40 seconds in this case). Plugging in the values, we get:
h = (1/2) * 9.8 * (40^2)
h = 0.5 * 9.8 * 1600
h = 7840 meters
So, the height of the jet is 7840 meters.
b. Since there is no horizontal force acting on the parcel during its free fall, its velocity in the horizontal direction remains unchanged. Therefore, the horizontal distance covered by the parcel during the 40 seconds will be equal to the product of the horizontal velocity of the jet and the time taken:
Distance = Velocity * Time
Distance = 500 m/s * 40 s
Distance = 20,000 meters
Therefore, the parcel moves forward horizontally by 20,000 meters.
c. The parcel falls vertically downwards due to gravity, while the jet continues to move forward with a constant velocity. Since there is no air resistance, the horizontal and vertical motions of the parcel and the jet are independent of each other. So, the parcel will land vertically below the position where it was dropped from the jet. Consequently, the parcel will be behind the jet by the horizontal distance traveled by the jet during the 40 seconds.
Therefore, the parcel is behind the jet by a distance of 20,000 meters.