A person in the passenger basket of a hot-air balloon throws a...

Question

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed 12.0 m/s (Fig. 3–64). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground
(b) if the hot-air balloon is descending at 3.0 m/s relative to the ground?
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(b) if the hot-air balloon is descending at 3.0 m/s relative to the ground?

Accepted Answer

Welcome back. Everyone in this problem, an open lift is descending at a steady speed of 2.5 m per second relative to the ground. As shown in the figure below. A person throws the lift, a person in the lift, sorry throws a small box horizontally outward from the lift with a speed of 10 m per second. We want to determine the magnitude and direction of the initial velocity of the box relative to a person standing on the ground. For our answer choices A says it's 10.3 m per 2nd, 14 degrees below the positive X axis B says it's 10.3 m per 2nd, 76 degrees below the positive X axis C says it's 7.5 m per 2nd, 14 degrees above the positive X axis. And D says it's 7.5 m per 2nd, 76 degrees above the positive X axis. Now, let's think about what's going on here. OK. So we have our box that is, is is thrown outward from the lift. So if we were to represent that, I mean, we have a bit of a diagram here, but we can draw a simpler diagram to represent the box. OK. That is thrown outward horizontally with a speed of 10 m per second. And it's descending at a steady speed of 2.5 m per second. OK. And we are trying to figure out the magnitude and direction of the initial velocity relative to a person standing on the ground. Now, let's think about it, even though it's done horizontally with a speed of 10 m per second, it won't be moving perfectly horizontally because it's descending at a given velocity. OK. So our box is actually gonna be moving somewhat in this direction that's in our red represented by a red vector here. So this will be our initial velocity vo Now what else do we know here? What, what, what other information do we have? Well, if we think about it, OK. If we want to figure out our resultant or rather if we want to figure out our initial velocity that has a horizontal and vertical component, then we should be able to find that by finding its resultant. OK, we can find the resultant of our two vectors here. Now recall that the resultant of a vector in this case vo OK. The magnitude of our result and V is going to be equal to the square root of it if it's X component squared plus it's Y component squared. But here's the great news, we have its X and Y components because we know it's initially thrown out horizontally at 10 m per second and it descending at a steady speed of 2.5 m per second. So that would be the, the y component of the initial velocity. And let's make that negative because we know it's descending. So since we have both of those values, that means we can solve for the magnitude of our initial velocity because no, that means its magnitude is going to be equal to the square root of 10 m per second squared plus negative 2.5 m per second squared. Now, when we put this expression into our calculator, we should find that we get the magnitude to be equal to 10.308 m per second, which to the nearest 10th of a meter per second is equal to 10.3 m per second. So that's the magnitude of our initial velocity. Next, let's find its direction. If we go back to our diagram, notice here that we could impose the Y, the Y component here. OK. And now we have an angle to the positive Y axis below the, below the positive X axis, sorry below the positive X axis that would represent the direction of our vector. So what do we know about the direction of victors? Well, again, we can recall, OK, that the anger that vector mix or its direction is going to be equal to the inverse tangent of the magnitude of its Y component divided by its X component. In this case, we're talking about the components of the velocity. So if we substitute those values, then we're going to be finding the inverse tangent of the Y component negative 2.5 m per second divided by the X component, 10 m per second. Now, if we put again, if we put this expression into our calculator, we should find that we get our anger to be 14 degrees. And again, where is it? Well, notice here that if we think about our axes this way, let me put it in blue, then it's below the positive X axis below the positive X axis. So that 14 degrees would be where we see theta in this diagram. Therefore, the magnitude and direction of our initial velocity is 10.3 m per second at 14 degrees below the positive X axis. When we look back on our answer, choices, we realize the correct answer is choice. A thanks a lot for watching everyone. I hope this video helped.

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