Problem

A puck is moving on an air hockey table. Relative…

04:05

Problem 15 Medium Difficulty

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Answer

Thax light $=2.8 \mathrm{m}$
Distance $=2 \mathrm{m}$

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Physics 101 Mechanics

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Rutgers, The State University of New Jersey

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Video Transcript

So the question states that a skateboarder is launched off of 58 degree angled ramp at 6.6 meters per second and at the edge of the ram, the height is 1.2 meters and we're trying to find where the skaters highest point is as well as how far in the X direction. Um, this highest point is located at, so the first thing really we should do is ah, break our velocity into components. So we know our V of X velocity component is going to be 6.6 times the co sign of this angle 58 degrees. And we know that our via why is gonna be equal to 6.6 times the sign of 58 degrees. And we know this because we know the sign of 58 degrees is equal to via y over a 6.6. So opposite overhype on you. So we know the coastline of 58 degrees is equal to via Vex that Jason over 6.6. So now that we know this, we can use our cinematics equations to solve for each of these problems. Um, for the first problem, we're trying to find the highest point we can use became an equation that states that the final velocity V squared is equal to the initial velocity square plus to a Delta X. And so we know when the skater that its highest point, his vertical velocity, is going to be zero. Um, so we can say that this will be equal to zero. We know his initial velocity is equal to 6.6 sign of 58 degrees in the Y direction squared. And then we know that the acceleration is going to be equal to negative 9.8 because of gravity and we're trying to solve for Delta X, which is the displacement from the top of this ramp to his highest point. And so when we saw for Delta X, so for tell to eggs, we find that tell tax is equal to 1.6 meters. But because we're doing this with the origin at the top of the ramp, we need to add 1.2 meters onto this because, um, this is not included in our calculation. So at 1.2 meters and we'll get that the final answer is 2.8 meters to find how far in the X direction Um, he's traveled. We really need to find the time it takes to get to this high position. So to do this, we can use our can amass equation that states that VI equals vi not plus a T. We know that the vertical velocity at this maximum point is going to be equal to zero again. And we know the initial velocity is 6.6 sign of 58 degrees. And we know the acceleration is gonna be negative 9.8 and then times that time so we can solve for T, where t is equal to 0.571 seconds. And now that we know the time, we can just use, uh, the Kid Max equation that says Delta X is equal to be not to plus 1/2 a t squared. And there's no acceleration in the X direction so we can get rid of this term. And we know the initial horizontal velocity is 6.6, uh, co sign 58 degrees this multiplied by the time is going to give us the distance. He's traveled from this, uh, origin point to this, Max. And we know the time because we calculated up here so complex that in when we do this, we get Delta X is equal to two meters.

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