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As preparation for this problem, review Conceptual Example 10. The two stones described there have identical initial speeds of $v_{0}$ 13.0 m/s and are thrown at an angle 30.0, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?

$\Delta x=14.935 \mathrm{m}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Hope College

University of Sheffield

University of Winnipeg

McMaster University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

10:12

A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.

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As preparation for this pr…

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Two stones are thrown up s…

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Two stones are thrown simu…

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A stone is thrown horizont…

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A stone thrown off a bridg…

04:30

\Lambdat what point on a p…

07:22

A stone is catapulted at t…

06:15

Stone $\mathrm{A}$ is thr…

02:50

A stone is thrown with ini…

02:15

A stone is thrown straight…

So the question states that two rocks are shot at 13 meters per second, 1 30 degrees above the horizontal and 1 30 degrees below the horizontal. And we're trying to find the distance between the two rocks when they hit the ground. So the first thing to note is that the rock that shot 30 degrees above the horizontal is going to fly up in the air, reach a peak up here and then cross back over where it started. But some distance, let's call it X away. And so at this point, when it's in the line from where it was shot, the angle of its velocity is going to be the same as the angle of the velocity of 30 degrees below the horizontal. It's also gonna be 13 meters per second as well, going below the horizontal, so it will just be an identical shot to this one. So this distance between them is going to stay constant as, uh, the both fall because they have the same projectile motion. So all we really need to do is figure out what this distance is. So how far the rock above the horizontal travels in some time T and that'll tell us how far away these two rocks are from each other when they hit the ground. So to do this, we can use our cinematics equations. Um, the 1st 1 we should look at is the one that states that the change in displacement is equal to the initial velocity times the time plus 1/2 times the acceleration times the time squared. Um and so we should focus on the vertical component of the velocity for this case because this will give us the time. It takes the the rock to cross back over this horizontal here in that we've labeled X. So the initial vertical velocity of the rock above the horizontal we can use trigonometry to find that. So we know that the sign of 30 degrees should be equal to the opposite, which will call V supply the vertical component of the lost E overhead part news, which means V sub Y is equal to 13 sign of 30 degrees. So this is gonna be RV not in the equation up here. We know that Delta X should be zero because it crossed. It starts on this plane and ends on this plane that we've labeled X. So why have zero year is equal to 13 sign of 30 degrees times time plus 1/2 times the acceleration, which is due to grab a piece of negative 9.8 times Times Square and so we can factor at a T. And so, for, um, the time that it takes so should be zero is equal to 13 sign 30 degrees plus 1/2 times negative 9.8 t in all this times t So we'll find that the roots for time are either t equals zero, which doesn't make any sense or tea equals to 1.32 So we need to use this route. The time is in zero. So now that we know the time that the project I was in the air, we can just use our equation for the range which states that the initial velocity in the horizontal direction times the time should be equal to the range of the projectile. So, like we did, uh, up above. For this, we can find the initial horizontal velocity of the project because co sign of 30 degrees his same thing as adjacent overhype on news Jason. We're just calling Lisa backs, and the hype on news is 13 meters per second. So that means Visa of X is equal to 13 co sign of 30 degrees. And now that we know this, we can plug it into the equation appear and we also know the time from appear and we complete this in as well, and we'll find that Delta X should be equal to 14 0.935 liters, and that is our answer.

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