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University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

00:56

Donald Albin

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

01:23

Suman Saurav Thakur

(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 $g$ of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

04:39

Muhammed Shafi

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

0:00

Aditya Panjiyar

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welcome to our second example video on projectile motion where we'll consider a problem that's often seen in the movies where you have a vehicle of some sort that's driving off a cliff. Okay, so in this case, we're gonna pick some generic variable names, perhaps will substitute them later with some numbers will say that we start at a height H. And what we want to know is, how far are we going to travel during our time of flight here? So we have some initial velocity that I'm just gonna call Vienna. All in the I had direction. So Vienna is how fast removing and the direction is 100% in the eye hat direction. Meanwhile, our initial position is that why not equals H? We'll go ahead and call this The X equals zero line here at the edge of the cliff. So when we pass over X equals zero, we leave the cliff with a velocity of Vienna. I had and at a height of H. So, in order to solve this, we're gonna use the same equations Last time where we can say that I know that my exposition it's going to be equal to Vienna. X Times time. Well, my wife position. Remember, generically, we have an equation that looks something like this. We need to think, Do I know my acceleration? And yes, we do. It's going to be equal to gravity. So that's negative. Little G where little G is 9.8 m per second squared V not. Why, however, is zero. So we're gonna end up with y equals H minus one half g t squared for our equation that predicts where we'll be in the Y direction now. Our final position that were curious about is when y equals zero. So let's go ahead and plug that in zero equals H minus one half g t squared. And if we solve for the time here, it will tell us. At what time does the car arrive at the bottom of the cliff were given a time here. We can go ahead and check the units. We have units of meters divided by meters per second squared. We're taking the whole thing to the one half power. So meters cancels second squares, ends up in the numerator and then is rooted, so we end up with just seconds. So this is good calculation here. Then we can take our time of flight, the time to go from the top of the cliff to the bottom of the cliff and plug it into our equation for the Exposition X is equal to V, not ex times the square root, then of two h over G and having plugged that in, we have a correct answer. We can again check the units, find that we have meters per second times seconds. The seconds cancel and we're left with just meters, which is exactly what we wanted was a position. So this is fairly quick problem to solve, especially since we previously solve the range equation. Now notice that we cannot use the range equation in this case because we're not ending at the same Y position that we started at. Therefore, it's not valid to use the range equation, but still there's not a significant amount of arithmetic that has happened. Once you've set up your two equations here, it only took a couple steps in order to come to the final solution. Always be looking for how to set up your equation first and then start worrying about the algebra If you start worrying about the algebra too soon or start plugging in numbers too soon, you're almost guaranteed to get lost. In this case, I could go through and I could set numbers like the height of the cliff was 50 m and the initial velocity Waas 12 m per second or 22 m per second. Something like that. And negative g, of course, is negative 9.8 m per second squared and then I could go in and plug all these numbers in and calculate where the object will land. Um, if you'd like to do that yourself, I definitely encourage you. There's almost certainly an example like this in your book.

Newton's Laws of Motion

Applying Newton's Laws

Work

Kinetic Energy

Potential Energy

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