Question

Your starship, the aimless wanderer, lands on the mysterious planet mongo. As the chief scientist-engineer, you make the following measurements: a 2.50 kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 5.70 s; the circumference of mongo at the equator is 2.00×105 km; and there is no appreciable atmosphere on mongo. What is the height reached by the stone?

Answer 1

The stone thrown upward at 12.0 m/s on planet Mongo reaches a maximum height of 17.1 meters, calculated using kinematics and the measured time of flight.

Step-by-step explanation:

To calculate the maximum height reached by the stone on the planet Mongo, we can use the principles of kinematics, specifically the equations of motion for uniformly accelerated motion (assuming gravity is the only acceleration affecting the stone). Since there is no atmosphere to create air resistance, this simplification is reasonable. The stone takes 5.70 seconds to return to the ground after being thrown upwards at 12.0 m/s, and it will reach its maximum height at the halfway point in time, which is 2.85 seconds.

We can use the following kinematic equation:

• h = v0t + (1/2)at2

Where:

• h is the height reached
• v0 is the initial velocity (12.0 m/s)
• t is the time to reach the maximum height (2.85 s)
• a is the acceleration due to gravity, which we will need to calculate using the entire duration of the motion (5.70 s)

First, we find the acceleration due to gravity (g) on Mongo by recognizing that the stone's upward motion and downward motion are symmetrical and using the total time of flight (5.70 s) to determine it:

g = (2 * v0) / t

g = (2 * 12.0 m/s) / 5.70 s

g = 4.21 m/s2

Now, we can find the maximum height reached by the stone:

h = 12.0 m/s * 2.85 s - (1/2) * 4.21 m/s2 * (2.85 s)2

h = 34.2 m - (1/2) * 4.21 m/s2 * 8.1225 s2

h = 34.2 m - 17.1 m

h = 17.1 m

The stone reached a maximum height of 17.1 meters on planet Mongo.