Posted by **JC** on Sunday, September 10, 2006 at 2:12pm.

A tennis ball on Mars, where the acceleration due to gravity is 0.379 of a and air resistance is negligible, is hit directly upward and returns to the same level 9.50 later.

How fast was it moving just after being hit?

The formula is the same as we use on earth, but the constants are different.

v(t)=-gt+v where g=1/3 earth's gravity.

Is that 9.5 sec?

If it takes the same time to go up that it does to come down, then solve

0=-g*4.25+v where g=.379a

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