An astronaut on Mars kicks a soccer ball at an angle of 45.° with an initial velocity of 15. m/s. If the acceleration of gravity on Mars is 3.7 m/s, a) What is the range of the soccer kick on a flat surface? b) What would be the range of the same kick on the Moon, where gravity is one–sixth that of Earth?

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:

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