Answer:
It will reach 348 m.
Explanation:
The gravity acceleration on Earth is
[tex]g=9.81\frac{m}{s^{2}}[/tex]
And for the Moon we write :
[tex]g_{Moon}=\frac{g}{6}[/tex]
Being g the gravity acceleration on Earth
The speed equation on a vertical throw is the following :
[tex]V_{f} ^{2}=V_{0} ^{2}-2g(y-y_{0})[/tex]
Where [tex]V_{f}[/tex] is the final speed of the throw
[tex]V_{0}[/tex] is the initial speed of the throw
g is the acceleration (generally the gravity acceleration)
y is the height reached
And [tex]y_{0}[/tex] is the initial height
We define [tex]y_{0}=0m[/tex] putting the comparison plane on the Earth surface.
Applying the equation on Earth we have the following :
[tex]0=V_{0} ^{2}-2g(58m)[/tex]
Because the final speed is 0 when the object is in it maximum height
and 58 m is the difference between the maximum height and [tex]y_{0}[/tex]
Then ⇒
[tex]V_{0}^{2}=2g(58m)[/tex]
Finally,applying the equation on the Moon :
[tex]0=V_{0}^{2}-2\frac{g}{6}(y-y_{0})[/tex]
We use the expression of [tex]V_{0} ^{2}[/tex] that we calculated :
[tex]0=2g(58m)-2\frac{g}{6}(y-y_{0})\\2\frac{g}{6}(y-y_{0})=2g(58m)[/tex]
[tex]\frac{(y-y_{0})}{6}=58m[/tex]
[tex]y-y_{0}=348m[/tex]
Then 348 m would be the height variation between [tex]y_{0}[/tex] and y
If we define [tex]y_{0}=0m[/tex] putting the comparison plane on the Moon surface.
The height reached will be 348 m.
