A 4.03-kg object is thrown vertically upward from the surface of Earth, where the acceleration due to gravity is g1 = 9.81 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 = 1.27v1 from the surface of Saturn\'s moon Tethys? The acceleration due to gravity on Tethys is g2 = 0.145 m/s2. Give your answer as a multiple of y1.
To find the maximum height, y2, on Tethys, we can use the principle of conservation of energy.
The total mechanical energy of the system remains constant throughout the motion. We can express the total mechanical energy as the sum of kinetic energy (KE) and potential energy (PE):
E = KE + PE
For the object thrown vertically upward from the surface of Earth, the kinetic energy at the maximum height is zero since the velocity becomes zero. Therefore, we can set the total mechanical energy on Earth equal to the potential energy at the maximum height:
E1 = PE1 = mgh1
Where m is the mass of the object, g1 is the acceleration due to gravity on Earth, and h1 is the maximum height on Earth.
On Saturn's moon Tethys, the object is thrown with a speed of v2 = 1.27v1. The kinetic energy at the maximum height is again zero, so we set the total mechanical energy on Tethys equal to the potential energy at the maximum height:
E2 = PE2 = mgh2
Where g2 is the acceleration due to gravity on Tethys, and h2 is the maximum height on Tethys.
Since the total mechanical energy is conserved:
E1 = E2
mgh1 = mgh2
Canceling out the mass on both sides:
gh1 = gh2
Dividing by g1:
h1 = h2/g1
Now, we can substitute the given values to calculate the maximum height on Tethys, y2, as a multiple of y1:
y2 = h2 = h1 * (g2/g1)
= y1 * (g2/g1)
Therefore, the maximum height, y2, on Tethys is y2 = y1 * (g2/g1).
To find the maximum height, y2, we first need to understand the relationship between the initial velocity, acceleration due to gravity, and maximum height.
In general, we can use the following equation to determine the maximum height reached by an object thrown vertically upward:
y = (v^2 - u^2) / (2g),
where:
- y is the maximum height,
- v is the final velocity of the object at the maximum height,
- u is the initial velocity of the object,
- g is the acceleration due to gravity.
Given that the initial velocity is v1 and the acceleration due to gravity on Earth is g1, we can use these values to find y1:
y1 = (v1^2 - 0^2) / (2 * g1)
= v1^2 / (2 * g1)
Now, we are given that the object is thrown from the surface of Saturn's moon Tethys with a speed of v2 = 1.27v1 and the acceleration due to gravity on Tethys is g2 = 0.145 m/s^2.
To find the maximum height, y2, on Tethys, we can use the same equation:
y2 = (v2^2 - u^2) / (2 * g2)
= (1.27v1)^2 / (2 * g2)
= 1.61(v1^2) / g2
Now, we can express y2 as a multiple of y1 by dividing y2 by y1:
y2 / y1 = (1.61(v1^2) / g2) / (v1^2 / (2 * g1))
= 1.61 * 2 * g1 / g2
= 3.22 * g1 / g2
So, the maximum height, y2, on Saturn's moon Tethys is 3.22 times the maximum height, y1, on Earth.
Therefore, y2 = 3.22 * y1.
Situation 1:
u = v1
v = 0 m/s
a = -g = -9.81m/s^2
S = y1
use S = ut+(1/2)*at^2 to get y1 in terms of v1
Situation 2:
u = 1.27v1
v = 0 m/s
a = -g(saturn) = -0.145m/s^2
S = y2
use S = ut+(1/2)*at^2 to get y2 in terms of v1
Then use the two answers together to get y2 in terms of y1