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.

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Create a have program that has the output given in the attached photo

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
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
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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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