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Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.00 $\mathrm{m}$ . The stones are thrown with the same speed of 9.00 $\mathrm{m} / \mathrm{s}$ . Find the location (above the base of the cliff) of the point where the stones cross paths.

$y_{1}=2.455 \mathrm{m}$

04:20

Patrick C.

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

University of Washington

Simon Fraser University

Hope College

McMaster University

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for this problem on the topic of kinda Matics in one dimension, we are told that simultaneously two stones are thrown one straight upward from the base of a cliff and the other straight downward from the top of the same cliff If the height of the cliff is six m and the stones are both thrown with the speed of nine m/s. You want to find the location of the point where the stones will cross paths. Now the initial velocity v not is known for both stones as the acceleration a due to gravity. In addition, we know that at the crossing point the stones are at the same place at the same time. T furthermore the position of each stone is specified by its displacement. Why from its starting point. And so the equation of chromatic that relates the variables in question is why is equal to Vienna T plus a half a T squared. So we can apply this equation to each stone, we have the displacement for the stone moving upward, Y up is equal to we not up times T. Plus half a T squid. And for the stone moving downward Y. D. Is equal to the initial speed of the downward moving stone. Vrt times T plus a half a t squared in these expressions. T. Is the time it takes for either stone to reach the crossing point, and is the acceleration due to gravity. Note that Why Up is the displacement of the upward moving stone above the base of the cliff? Whereas why down is the displacement of the downward moving stone below the top of the cliff? Each year is the displacement of the clip cliff top above the base of the cliff. As we can see in the drawing now the distances above and below the crossing point must and to equal the height of the cliff. So why up minus Why? Down must equal to H where the minus sign appears because the displacement Y. Down points in the negative direction. If we substitute the two expressions for Why? Up and lie down into this equation. We get that. We not up times T plus a half a T squared minus. We not G. Times T plus a half 80 squared is equal two H. And then we can solve this equation for tea to show that the travel time to the crossing point T. Is H divided by. We not up -3 down. If we subject this into the equation of motion, we get that. The displacement why up is equal to. We're not up times T plus a half a. T squared. And this is we're not upwards into the expression for the time which is age over we, not up minus we, Not down plus a half A into H divided by we not up minus. We note down all squared. And so now if we substitute our values in, we get this to be nine m/s Into the height of the cliff, which is six m divided by nine m/s -9 m/s plus a half times the acceleration which is downward and hence negative 9.8 m the square second. We suppressed the units here, all multiplied by six m, divided by nine m per second minus minus nine m/s. And so calculating, we get this distance to be 2.46 m above the base of the cliff.

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