You are fixing the roof of your house when a hammer breaks loose ...

Question

You are fixing the roof of your house when a hammer breaks loose and slides down. The roof makes an angle of 35° with the horizontal, and the hammer is moving at 4.5 m/s when it reaches the edge. What are the horizontal and vertical components of the hammer's velocity just as it leaves the roof?

Accepted Answer

Everyone in this particular practice problem, we are asked to calculate the X and A Y component of the velocity factor at the instant when the car goes off the handrail, where we'll have a toy car sliding on the handrail of a house and handrail is 40 degrees incline with respect to the horizontal. At the end of the handrail, the car has a speed of 4.5 m per second and the car is going down the handrail. So in this particular example, I'm going to start with um drawing our system. So we will have the incline or the handrail here, width or making an angle of theta equals 40 degrees the theta being here and we'll have a car, I'm just gonna indicate the car with a cell circle right now. And the car will have aoc Of four or speed of 4. m/s, which is v equals 4.5 m/s. And next, we will of course have the X and Y axis which will make our projection easier. So I'm just gonna draw our X and Y axis first. So this is going to be our X axis, this is going to be our Y axis and we'll have the X and Y component of our V of our velocity. This is our P X and this is going to be our P Y. And we're being asked, what is the value or what's the magnitude of the X and Y components of those velocity factors or what is the X and Y is P Y. So what we want to indicate next is the uh actually what this angle is. And we can find what this angle is By actually looking into our known incline slope, which is 40° and with the properties of the angles being back to back to one another with the same two lines because this velocity is parallel to the incline and the uh V X is on the X axis along with the ground. So two of the same lines crossing one another, the back to back angle will be the same and that is a property off the angle. So when the car actually goes off the handrail, the velocity factor is directed at 40 degrees below 40° below the horizontal. So the x is pointing to the right and P Y is pointing downwards. So P X is going to be positive, pointing to the right and P Y is going to be pointing downwards. So P Y is going to be negative. So we will apply the trigonometric ratio and solve for V X and V Y. So the um triangle that we're gonna be working with is this triangle right here. And I'm gonna draw that real quick to make it easier for us to see. So this will be our triangle right here with a right angle being here. And this is going to be our V X and this is going to be our hypotenuse or V and this uh site right here is going to be our V Y with a theta of here. OK. So this is the triangle that we're gonna be working with. So we wanna recall that using this triangle right here, um P X can be calculated by equals P multiplied by a cosine of theta. And the V is going to be known from the problem statement which is 4.5 m per second and the theta is 40 degrees. So the theta is cosine of 40 degrees and this will give us the V X value of 3.4 m per second. And the V X is going to be positive because it's pointing to the right next, we want to calculate P Y which is V multiplied by sine of theta. And we want to recall that P Y is negative because it's pointing downwards. So I'm gonna multiply this with, with minus one. So next PY will be -4.5 m per second multiplied by sine of 40°. And that will give us AVY value of -2. m/s. So with A V X of 3.4 m per second and A V Y of minus 2.9 m per second, the answer to this problem is going to be option B so option B is going to be the answer to this problem and that will be all for this particular practice problem. If you guys have any sort of confusion, please still make sure to check out our other lesson videos on similar topics and that'll be it. Thank you.

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You are fixing the roof of your house when a hammer breaks loose and slides down. The roof makes an angle of 35° with the horizontal, and the hammer is moving at 4.5 m/s when it reaches the edge. What are the horizontal and vertical components of the hammer's velocity just as it leaves the roof?

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