Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of 45° with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of 30° with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. Ignore air resistance and the mass of the vine.

Question

Accepted Answer

Hey everyone. Welcome back in this problem. A certain artist needs to swing across a river using a 15 m long rope attached to a bridge. Initially, the rope makes an angle of 60 degrees with the vertical and the artist swings down before rising to meet a colleague on the bank of the river. When he reaches the other bank, the rope makes a 25 degree angle with the vertical and were asked to calculate the artist speed just before he releases the rope and is caught by his colleague. Alright, so let's just draw a little diagram here. So initially what do we have? We have the rope attached to the bank or to the bridge sir on the left hand side of the bank and we have our cirque artists holding onto that rope. The length of the rope is 15 m and it makes an angle 60° with the vertical. Alright, so we have our angle of 60° and this person is going to swing to the other side of the bank and when they get to the other side of the bank, they're gonna be letting go of that rope and we want to know the speed just before they let go of that rope. Okay. The rope is still 15 m. The rope doesn't change But the angle with the vertical now is 25°. So I'm just gonna write this out here 25° instead of 60°. Now we're gonna take downwards as our positive y direction and we're going to consider the, the height of the bridge where the rope is attached is going to be our reference point Where the height H is considered to be equal to zero. Alright, so we want to figure out the artist speed, okay. But we have two things to consider this artist is moving. So they have some kind of kinetic energy and then they're also at some height. So they have potential energy due to gravity as well. So let's consider the conservation of mechanical energy. We can take both of those things into consideration. So we have the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. Alright. Well, the initial kinetic energy is going to be 1/ M the knot squared. In our potential energy, we just have gravitational potential energy that's going to be equal to M G times H not that initial height of our cirque artist in the final kinetic energy one half M the F squared, the final potential energy is going to be due to gravity. So gravitational potential energy M G H F. And what we're looking for is the artist speed just before he releases the rope, which is going to be that final speed V F. You'll notice that in every term here, we have the quantity M, the mass is the same. The cirque artist doesn't change mass through this swing. And so we can divide by M from the left and right hand side and that is going to go away in each of those terms. No, initially, the artist is at rest, he's holding the rope, he's at rest and then he starts to swing. So his initial speed V not is going to be zero. So initially, we're just gonna have the gravitational acceleration times, the initial height Is equal to 1/2 V F squared plus G H F. And we want to solve for V F. So let's go ahead and do that. Let's isolate for VF squared V F squared is going to be equal to, we're gonna have to subtract this G H F term and then we're gonna have to multiply by two. And so we get two times G H not minus H F. Alright. So we know the gravitational acceleration G, we need to figure out a church, not an H F in order to determine R V F squared. All right, let's do it. So let's think about our triangles here. We're going to need to do some triangle math. So initially, we have the following triangle. We have the hypothesis, 15 m, the length of the rope, we have the angle With the vertical of 60°. And we want to figure out the height age. Not alright. So each not that's the adjacent side. So this is gonna be equal to coast of 60° times 15 m. And if we do the same for our final triangle, We have a similar triangle. It has a hypotenuse of 15 m, the length of the room, An angle with the vertical of 25°. We want to know the height H F And again, HF because we are talking about the adjacent side, it's going to be related through coast to coast of 25° times 15 m. All right. So we have our two heights here. We're talking about these heights, we're talking about the magnitude of the height. So when we get into our equation, we're going to take the absolute value because we're talking about again, just that magnitude of their height. And so V F squared Is going to be equal to two times gravitational acceleration 9.8 m per second squared. And then we're gonna take the absolute value of co sign 60 degrees times 15 m minus cosine of 25 degrees times m. This is gonna give us a V F squared value of 119 0. five m squared per second squared. And we had meters per second squared, then we multiplied by meters. So we have meters squared per second squared. When we take the square root, we're going to get a positive and a negative square root And we get 10 . m per second. And so the speed of our search artist is going to be the absolute value of this. It's going to be 10.93 m per second. So let me just write that out. So we aren't confused. The speed is just going to be the magnitude of that velocity Which is going to be the positive 10.93 m/s. All right, there you have it. If we go back up to our answer choice, Okay. We found that the artist speed just before he releases the rope is approximately 10.9 m/s. That corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Conservation of Energy

Kinematics of Circular Motion

Forces and Acceleration