2. The equation for the position of equivalent-speed kinetic objects is as follows. 1 ¤(t) = at + vot + ¤0 where x(1) is the position, a is acceleration, t is travel time, vo is initial velocity, and x, is the initial position. Assume that you try to drop an object from a standstill at an altitude of 1,000 meters. Complete the Java code and obtain the same result as the output example. (There is no resistance in free fall, and the acceleration of gravity is -9.81m/s.
2. The equation for the position of equivalent-speed kinetic objects is as follows. 1 ¤(t) = at + vot + ¤0 where x(1) is the position, a is acceleration, t is travel time, vo is initial velocity, and x, is the initial position. Assume that you try to drop an object from a standstill at an altitude of 1,000 meters. Complete the Java code and obtain the same result as the output example. (There is no resistance in free fall, and the acceleration of gravity is -9.81m/s.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
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Transcribed Image Text:2. The equation for the position of equivalent-speed kinetic objects is as follows.
1
æ(t) = at + vot + æo
where x() is the position, a is acceleration, t is travel time, vo is initial velocity,
and xo is the initial position. Assume that you try to drop an object from a
standstill at an altitude of 1,000 meters. Complete the Java code and obtain the
same result as the output example. (There is no resistance in free fall, and the
acceleration of gravity is -9.81m/s.
Position after 5.00 seconds: 877.38m
Figure 2 Output of the Second Program
Expert Solution
Step 1
Objective: The position of a kinetic object should be calculated after a given time when the acceleration due to gravity, time, initial position are provided. Also, the required outcome of the program is mentioned.
Programming language: Java
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Solved in 3 steps with 1 images
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