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While standing on a bridge 15.0 $\mathrm{m}$ above the ground, you drop a stone from rest. When the stone has fallen $3.20 \mathrm{m},$ you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

$v_{0}=-11.31 \mathrm{m} \cdot \mathrm{s}^{-1}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

Rutgers, The State University of New Jersey

University of Washington

University of Winnipeg

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can. This problem? One stone is dropped from the top of a bridge or from a 15 liter Highbridge with an initial velocity of zero when it has fallen the distance of 3.2 meters. A second stone is thrown from the same bridge, and the question is, how fast doesn't need to be thrown in order to hit the ground at the same time. So first, let's figure out how long it takes the first ball to fall 3.2 meters. They state that the problem should be negative in the downward direction. Some to do. Negative 3.2 leaders is the initial velocity times teat. Well, this one was dropped, so the initial velocity zero, So I didn't have to write it down. My acceleration is negative. 9.80 cause we're on the surface of the Earth Times t squared. So my negatives cancel my t squared comes out to be 0.653 2nd squared and taking the square root of that I get t is equal 2.808 seconds. So that's how long it takes from the ball to go from the bridge to the 3.2 mark. What would it take the first ball to go the whole distance? Because then, if we know how long it takes to go from here to the water surface, we know how long the second vault has, because it's gonna have less time to follow them the first ball if it's going to catch up with it. So let's do the 15 meters negative 15 is This also still has the same initial velocity of zero. So negative 9.8 times t squared. My T squared in this case is 3.6 2nd squared and we take the square root of that and we get 1.7496 I have too many significant figures, but that's OK for now. Okay, so stone to the second stone has that much less time because the first stone would take 1.7496 seconds to get to the ground. But it has the second stone has 0.808 seconds less in order to do that. So let's find the time for the second stone, which would be the difference of these two numbers. And I have 0.942 seconds, rounding to three Sync fix at this point. So that's how long my blue stone on my my second stone has to hit the water surface. So I need to figure out what that initial velocity is. So let's just scroll up. So I have a little more space. I know it's still falling 15 meters. I know the time at this point and I just don't know the initial velocity. So we use the same equation and books. So I have 15 meters in the negative direction. I don't know my initial velocity, but I do know my time is 0.9 for two seconds plus 1/2 negative, 9.8 liters per second squared times 0.942 seconds and that quantity is squared so we can simplify a little bit. Negative. 15 meters is 0.942 times the initial velocity. This whole term right here is negative. 4.43 4.34 My apologies, Meters, We, um, about two. Both sides have negative 10.65 meters. Um, 10.66 I think it would be can 0.66 meters equals V not times 0.9 for two and then solving by dividing each side by 0.942 We get 11 point three, and that's in the negative direction, and that's our answer.

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