The acceleration due to gravity on the moon is about one-sixth its value on earth. If a baseball reaches a height of 66 m when thrown upward by someone on the earth, what height would it reach when thrown in the same way on the surface of the moon?

I found: ##396m##

We can use the relationship from kinemamatics:
##color(red)(v_f^2=v_i^2+2ad##
with:
##v_f=## final velocity which is zero at the top;
##v_i=## initial velocity (the same on Earth and on the Moon I suppose);
##a=## accelration of gravity (##=g## downwards);
##d=## distance (height in this case).
We have:
On Earth: ##0=v_i^2-2g*66##
so that ##v_i^2=2g*66##
On the Moon: ##0=v_i^2-2/6gd##
so that: ##v_i^2=2/6gd##
so considering the same values for ##v_i^2## we can write:
##2cancel(g)*66=2/6cancelgd##
##d=6*66=396m##

Posted in Uncategorized

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes:

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>