Finding g on a small airless planet Explorers...

Question

Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/sec. Because the acceleration of gravity at the planet’s surface was gₛ m/sec², the explorers expected the ball bearing to reach a height of s = 15t − (1/2)gₛt² m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gₛ?

Accepted Answer

Hello. In this video, we are going to be answering the following word problem. So we are told that a scientist is studying the motion of a drone on a distant moon, where the gravity is different. The drone is launched vertically upward with an initial velocity of 10 m per second. The formula for its height at any time T is given by H equal to 10T minus 12 G M T squared. This is where GM is the moon's gravitational acceleration in meters per second squared. If the drone reached its peak height 15 seconds after launch, what is the value of GM? So, the first thing we want to label in this problem is that we are given a function for the drone's height. The function for the drone's height is given to us as H is equal to 10T minus 12 GM multiplied by T squared. Now, we want to determine the value of GM, but we are also told that the drone reaches its peak height at 15 seconds. Now, what does it mean to have a peak height of a graph? Well, the peak height looks like the top of a mountain on a graph, and that means that we're the problem is telling us that we have a local maximum or a maximum height. Located at T is equal to 15 seconds. So since we are given a value of T relating to a local maximum, let's go ahead and take the derivative of the given height function. If we take the derivative of the given height function, that is going to give us the velocity function. And that derivative is going to be 10 minus GMT. Now, we want to set this equation equal to 0 in order to determine the values of the maximum height, or the time or the time that occurs for the maximum height. But we are given that information. We are told that the peak height occurs at 15 seconds. So, if we set the velocity equal to 0 and plug in 15 for T, we get 10 minus GM multiplied by 15 is equal to 0. Now, what we want to do is we want to solve for GM. So, we will go ahead and add GM multiplied by 15 to both sides of the equation. That will give us GM multiplied by 15 is equal to 10. And dividing both sides of this equation by 15 is going to give us GM is equal to 10 divided by 15, which further simplifies to 2 divided by 3 m per second squared, and this is going to be the solution to this problem. And that means that the overall answer is going to be A. So, I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Key Concepts

Kinematic Equations

Maximum Height in Projectile Motion

Solving Quadratic Equations

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