A 4.51-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 = 1.39v1 from the surface of the Moon? The acceleration due to gravity on the Moon is g2 = 1.62 m/s2. Give your answer as a multiple of y1.
To find the maximum height, y2, of the object when thrown from the surface of the Moon, you need to compare the acceleration due to gravity on the Moon, g2 = 1.62 m/s^2, with the acceleration due to gravity on Earth, g1 = 9.81 m/s^2.
Let's break down the steps to solve the problem:
Step 1: Find the initial velocity on the Moon, v2:
Given that the object is thrown with a speed of v2 = 1.39v1 from the surface of the Moon, we can find v2 by multiplying the initial velocity on Earth, v1, by 1.39.
v2 = 1.39v1
Step 2: Calculate the time it takes for the object to reach its maximum height on the Moon, t2:
The time it takes for the object to reach its maximum height depends on the initial velocity and acceleration due to gravity. We can use the equation:
v = u + at
Since the object is thrown vertically upward, the final velocity, v, at the maximum height is zero.
0 = v2 - g2t2
By rearranging the equation, we get:
t2 = v2 / g2
Step 3: Find the maximum height, y2:
The equation to calculate the maximum height reached by an object thrown with an initial velocity and acceleration due to gravity is:
y = ut - (1/2)gt^2
Since the final velocity at the maximum height is zero, the equation simplifies to:
y2 = v2t2 - (1/2)g2t2^2
Now that you have all the necessary components, plug in the values of v2 and t2:
y2 = (v2/g2) * v2 - (1/2)g2(v2/g2)^2
Simplifying this equation will give you the maximum height on the Moon, y2, as a multiple of y1.