A ball thrown upward near the surface of the Earth with a velocity of 50 m/s will come to rest about 5 seconds later. If the ball were thrown up with the same velocity on Planet X, after 5 seconds it would be still moving upwards at nearly 31 m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the Earth? Using the MathType functions, be sure to show your work

Using v = u + at where v = velocity of ball after 5 s on planet X = 31 m/s, u = initial velocity of ball on planet X = 50 m/s , a = acceleration due to gravity on planet X and t = 5 s

So, 31 = 50 - a × 5 = 50 - 5a

31 - 50 = 5a

-19 = 5a

a = -19/5 = -3.8 m/s²

So, the magnitude of a = 3.8 m/s²

So a/g = 3.8/9.8 = 19/49

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Lanuel Lanuel · Answer 2 · 2021-11-23T22:49:02+01:00

The fraction of the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth is 0.39.

Given the following data:

Time = 5 seconds

Initial velocity = 50 m/s

Final velocity = 30 m/s

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]

To determine what fraction is the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth:

First of all, we would calculate the acceleration due to gravity (g) on Planet X by using first equation of motion:

Mathematically, the first equation of motion is calculated by using the formula;

[tex]V = U-at[/tex]

Where:

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]31=50-a(5)\\\\5a =50-31\\\\5a=19\\\\a=\frac{19}{5}[/tex]

Acceleration, a = 3.8 [tex]m/s^2[/tex]

For the ratio:

[tex]Fraction = \frac{a}{g} \\\\Fraction = \frac{3.8}{9.8}[/tex]

Fraction = 0.39

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