(III) A falling stone takes 0.28 s to travel past a window 2.2...

Question

(III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2–49). From what height above the top of the window did the stone fall?

Accepted Answer

Hey, everyone in this problem, a ball falling from rest takes 0.64 seconds to pass the platform that is 4 m tall. We're asked to determine the height above the top of the platform from which the ball fell. We're given four answer choices all in meters. Option A 4.7 option B 0.49 option C 0.87 and option D 6.8. So let's start by drawing a little diagram of what we have going on. So we have this 4 m high platform right now. We're gonna have a ball that is dropped from some height above that platform. We're gonna call that distance between the ball and the top of the platform. H OK. And that is that value that we're looking for. That's the height we're looking for. Now, we're told that the ball falls from rest. And so the initial velocity of the ball is going to be 0 m per second now, because the balls falling downwards, let's take downwards to be our positive direction. That's gonna just make it a little bit simpler without having a lot of negative signs. OK. All right. So let's think about the ball falling from the initial position to this platform. OK. This distance of age. OK. So let's do this in blue. So this fall is going to be in blue. OK? So this is from drop to the platform. And again, we know that the initial speed or velocity is 0 m per second. We don't know what the final speed will be. We know that we have the acceleration due to gravity, which is gonna be 9.8 m per second squared. And that's a positive value because we're taking downwards to be our positive direction. Now delta Y that vertical displacement that's gonna be that value of H again that we're looking for. And finally the time this takes tea. Well, we don't know. Hm All right. So we want to calculate H but we don't actually have enough information. OK? We can see that we only have the not in a so in order to calculate, we need some more information. OK? So let's think about the rest of the motion. OK? Let's think about when the ball is at the platform, the top of the platform until the bottom of the platform. OK? We know that this takes 0.64 seconds. We were told that it takes 0.64 seconds to pass this 4 m platform. OK. So let's go from the platform, from the top of the platform to the bottom. All right. So the initial velocity V not. And we're gonna call this subscript P to indicate that we're talking about this platform portion of our motion. OK? We don't know the initial speed, OK? We don't know the speed that is the ball is going at the top of the platform, but we do know that this is equal to VF OK. The final speed if we're looking at the drop from the initial position to that top of the platform is the same as the initial speed if we're looking from the top of the platform down. OK. So these two are equal. All right. Now, the final velocity, we don't know either. We do know the acceleration again, the acceleration due to gravity 9.8 m per second squared. So delta Y in this case is going to be 4 m and we're following 4 m, that's the height of the platform. And we know that it takes 0.64 seconds. So in this case, we have three known values, we can calculate the others. OK? So what we're gonna do is we're gonna use the motion from the top of the platform to the bottom to calculate V not P pay the speed at the top of the platform, which we can use as our final speed when we're looking at this ball dropping from the top down to that platform. OK? Once we have that speed, we'll be able to calculate H All right. So starting with finding V not P, which we've said is equal to VF. Now, we want to use the equation with the three known variables we have as well as the one we're looking for being not. So we're gonna choose the kinematic equation. Delta YP is equal to V, not P multiplied by TP plus one half multiplied by A P multiplied by TP squared, going ahead and substituting in all the values we have, we get 4 m is equal to V. Not P multiplied by 0.64 seconds plus one half, multiplied by 9.8 m per second squared, multiplied by 0.64 seconds squared. Simplifying on the right hand side, 4 m is equal to V not P multiplied by 0.64 seconds plus 2.00704 m subtracting this 2.00704 m from both sides. We get that V not P multiplied by 0.64 seconds is going to be equal to 1.99296 meters. And we can divide by that 0.64 seconds on both sides to get that are the speed of the ball at the top of the platform is 3.114 m per second. All right. So now we know the speed at the top of the platform. Let's go back up to our values and to our diagram and we're gonna fill this in. Ok. So we know now that when the ball is at the top of the platform, the speed is 3.114 meters per second. And that's the initial speed when we're looking at this green section. Ok? When we're looking from the platform down, that's our initial speed, 3.114 m per second. And when we're looking at the other case, when we're looking at from where the ball drops to the top of the platform, that is our final speed. Mhm All right. So that gave us a lot of information, finding that one value. And now if we look at this blue information, OK. So from where the ball drops down to the top of the platform, we have three known values, we wanna calculate delta Y. And so we're gonna choose the kinematic equation with those four variables. All right. So now we are finding the height H which is really what the question was asking for. So the equation we're gonna use VF squared is equal to V not squared plus two multiplied by a multiplied by delta Y substituting in our values 3.114 m per second. All squared is equal to V not squared, which is just zero. And then we have two multiplied by the acceleration 9.8 m per second squared multiplied by the height H. So the height H is going to be two oops is going to be 3.114 m per second award divided by two, multiplied by 9.8 m per second squared. And we get a final height H of 0.4947 m. OK. All right. So that is the height we were looking for if we go back up to our diagram and this is the height that the ball fell above the platform. And we found that it is approximately 0.98 m which corresponds with answer choice. B thanks everyone for watching. I hope this video helped see you in the next one.

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(III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2–49). From what height above the top of the window did the stone fall?
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Struggling with Physics?

(III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2–49). From what height above the top of the window did the stone fall?<IMAGE>

Watch next

Vertical Motion and Free Fall practice set

(III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2–49). From what height above the top of the window did the stone fall?
<IMAGE>