Suppose that on earth you can throw a ball vertically upward a distance of 2.39 m. Given that the acceleration of gravity on Mars is 3.80 m/s2, how high could you throw a ball on Mars? (Take the y-axis in the vertical direction, and assume that the location of your hand is at y = 0.)

Answer 1

First, let's find out the initial velocity when you throw the ball vertically upward on Earth. We know the following information:

- Maximum height (h) on Earth: 2.39 m
- Acceleration due to gravity on Earth (g_e): 9.81 m/s^2
- Acceleration due to gravity on Mars (g_m): 3.80 m/s^2

We can use the following kinematic equation to find the initial velocity (v_initial) on Earth:

v_final^2 = v_initial^2 - 2 * g_e * h

Since v_final is 0 m/s at the top of the trajectory, the equation becomes:

0 = v_initial^2 - 2 * g_e * h

v_initial^2 = 2 * g_e * h

v_initial = sqrt(2 * g_e * h)

v_initial = sqrt(2 * 9.81 * 2.39) = 6.88 m/s

Now we know the initial velocity on Earth, let's find out the maximum height on Mars (h_m) with the same initial velocity:

v_final^2 = v_initial^2 - 2 * g_m * h_m

0 = 6.88^2 - 2 * 3.80 * h_m
h_m = (6.88^2) / (2 * 3.80)

h_m = 6.6402 m

So, you could throw a ball 6.64 meters high on Mars.