A 2.83-kg object is thrown vertically upward from the surface of Mars, where the acceleration due to gravity is g1 = 3.71 m/s2. The initial velocity is v1, and the object reaches a maximum height of y1. What is the maximum height, y2, if the object is thrown with a speed of v2 = 1.39v1 from the surface of Saturn\'s moon Rhea? The acceleration due to gravity on Rhea is g2 = 0.264 m/s2. Give your answer as a multiple of y1.

y2=_______* y1

consider energy:

beginning energy=final energy
1/2 m v1^2=m*3.71*y1
on saturn
1/2 m (1.39v1)^2=m*.264*y2

divide equation 2 by equation 1

1/1.39^2=.264/3.71 * y2/y1

y2=3.71/(.264*1.39^2) * y1

check my math

To solve this problem, we first need to find the maximum height, y2, in terms of y1.

The maximum height reached by the object can be determined using the conservation of mechanical energy principle. At the maximum height, all the initial kinetic energy of the object will be converted into potential energy.

Let's calculate the initial kinetic energy (KE1) and potential energy (PE1) of the object on Mars:

KE1 = (1/2) * mass * velocity^2
= (1/2) * 2.83 kg * (v1)^2
= 1.415 * (v1)^2

PE1 = mass * gravity * height
= 2.83 kg * 3.71 m/s^2 * y1
= 10.5293 * y1

According to the conservation of mechanical energy, KE1 + PE1 = PE2, where PE2 is the potential energy at the maximum height on Rhea.

Now, let's calculate the potential energy on Rhea (PE2):

PE2 = mass * gravity * height
= 2.83 kg * 0.264 m/s^2 * y2
= 0.75432 * y2

Setting PE2 equal to KE1 + PE1, we have:

0.75432 * y2 = 1.415 * (v1)^2 + 10.5293 * y1

Dividing both sides of the equation by y1, we get:

(0.75432/y1) * y2 = 1.415 * (v1)^2/y1 + 10.5293

Simplifying further, we have:

0.75432 * (y2/y1) = 1.415 * (v1)^2/y1 + 10.5293

Rearranging the equation, we find:

y2/y1 = (1.415 * (v1)^2/y1 + 10.5293) / 0.75432

Therefore, the maximum height, y2, is equal to (1.415 * (v1)^2/y1 + 10.5293) / 0.75432 times y1.

So, the answer to the question is:

y2 = (1.415 * (v1)^2/y1 + 10.5293) / 0.75432 * y1.