Answer:
60.7 meters on Mars.
137.8 meters in the Moon
Step-by-step explanation:
We have information about the problem statement, the following data:
Vi = 15 m / s
w = 45 °
Gmars = 3.7 m / s ^ 2
The first thing is to calculate the velocity components, like this:
Vix = 15 * cos (45) = 10.6 m / s
Viy = 15 * sin (45) = 10.6 m / s
The time t in the delay in reaching the maximum vertical displacement in y, is given by the following equation:
Vfy = Viy - Gmars * t
Knowing that Vfy is equal to 0, we can calculate the value of t:
0 = 10.6 - 3.7 * t
t = 10.6 / 3.7 = 2.87 seconds, but it should be borne in mind that the total time taken is double therefore it would be 2.87 * 2 = 5.73 seconds.
Now, the movement in x is given by the equation:
x = Vix * t
Replacing we have:
x = 10.6 * 5.73 = 60.7 meters.
60.7 meters would be the range of the kick on Mars.
Moon
Now, on the moon the same procedure is repeated, only the gravity that is 1/6 of the earth changes, therefore:
9.8 * (1/6) = 1.63 m / s ^ 2, now the equation would be as follows:
Vfy = Viy - Gmoon * t
Knowing that Vfy is equal to 0, we can calculate the value of t:
0 = 10.6 - 1.63 * t
t = 10.6 / 1.63 = 6.5 seconds, but it should be borne in mind that the total time taken is double therefore it would be 6.5 * 2 = 13 seconds.
x = Vix * t
Replacing we have:
x = 10.6 * 13 = 137.8 meters.
137.8 meters would be the range of the kick in the moon.
Step-by-step explanation:
Below is an  attachment containing the solution.
