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

To find the initial velocity of the tennis ball, we can use the equation for the motion of the ball after it has been hit:

v(t) = -gt + v

Where:
- v(t) is the velocity of the ball at time t
- g is the acceleration due to gravity on Mars (0.379 times the acceleration on Earth)
- v is the initial velocity of the ball

Since the ball returns to the same level after 9.50 seconds, we can assume that it takes the same amount of time to go up and come back down.

Let's calculate the time it takes for the ball to reach its peak height:

0 = -g*t_max + v

Where t_max is the time it takes for the ball to reach its peak height.

Since the time it takes for the ball to reach its peak is half the total time, we can substitute t_max with t/2:

0 = -g*(t/2) + v

0 = -g*t + 2v

So, we have two equations:
1) v(t) = -gt + v
2) 0 = -g*t + 2v

Substituting the value of g in equation (2) with 0.379a, we get:

0 = -0.379a*t + 2v

Rearranging equation (2), we can solve for v:

2v = 0.379a*t

v = (0.379a*t) / 2

v = 0.1895a*t

Now we have the initial velocity of the ball, which is v = 0.1895a*t.

Since we know that the total time is 9.50s, we can substitute t with 9.50s:

v = 0.1895a * 9.50s

v = 1.8a s

Therefore, the ball was moving at a speed of 1.8 times the acceleration due to gravity when it was just hit.