(II) Roger sees water balloons fall past his window. He notice...

Question

(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk.

(b) Assuming the balloons are being released from rest, from what height are they being released?

Accepted Answer

Everyone in this problem, a boy standing on a cliff, 30 m above the ground observes a bird in the sky dropping a stone, the stone hits the ground 1.58 seconds after it passes the cliff. And we're asked to determine the height from which the bird releases the stone. And we're gonna assume that the stone is initially at rest. And we're given four answer choices all in meters. Option A 36.4 option B 6.44 option C 11.2 and option D 46.1. So we're giving our diagram, we have our bird dropping the stone, OK? And then we have this cliff that is 30 m high, ok? And that stone is released from some height above. Ok. So the total height, the height from which the bird releases a stone that we were looking for is going to be the height from the stone all the way to the ground. Ok. And we're gonna call this delta Y now, if we think about this entire drop, ok, from where it's released until it hits the ground, what do we know? OK. Well, we know the stone is released from rest. So we knot is going to be 0 m per second. We aren't sure what the final velocity is. When the stone hits the ground. We have an acceleration. Ok? Let's say that downwards is going to be our positive direction. Then that acceleration is gonna be positive 9.81 m per second squared. OK? We wanna find delta Y and we don't know how long it takes for the stone to fall. And we know that it takes 1.58 seconds after it passes the cliff. But from the time it's dropped till the time it hits the ground. We don't know that time. And so we wanna find delta Y. And in order to do this, let's break it up into two pieces. OK? We don't have enough information here. We only have two known values. Remember, we need at least three in order to use our kinematic equations to find an unknown value. So let's break this up. OK. Again, we know that time if we're considering just after the cliff. So let's look at from the cliff height to the ground good. And when we do that, what do we have? OK. Well, we're gonna use subscript C to indicate this is from a clip. OK? We don't know the initial velo velocity V not see. OK. We don't know the speed when that stone is passing the cliff. OK. We're looking at this last 30 m when the stone is next to the cliff again, we don't know the final velocity when this thing hits the ground. We do know the acceleration. OK? I guess the acceleration due to gravity 9.81 m per second squared. We know Delta YC. Hey, delta YC is just gonna be the height of the cliff, which is 30 m and we know that the time this takes, which we'll call TC is 1.58. All right. So we do have three known values here. But we want to think about how is this going to help us find Delta War? OK. Well, we have one piece here and we have Delta YC, but we can also consider the first portion of this movement. OK? From when the stone is dropped until it reaches that cliff. OK. So from the drop to the cliff, we can consider this as delta Y and we're gonna call this BK before the cliff. All right. Now, in this case, we know the initial velocity is 0 m per second. The stone starts from rest. We don't know the final velocity, but we do know that it's gonna be equal to the initial velocity for, for our cliff to ground portion. OK. So the final velocity of the stone as it reaches the cliff is the same as the initial velocity if we're looking from the movement from the cliff on. OK. So these two are gonna be equal. So this allows us to relate these two different calculations. In these two different situations. We have our acceleration again due to gravity 9.81 m per second squared. We don't know what delta YB is and we don't know how long this takes. All right. So we have um quite a few unknown values. Now, again, let's not forget what we're looking for. We are looking for delta Y, the total height from the um where the stone is dropped. OK. Now, delta Y, as we can see from our diagram can be written as delta YV plus delta YC. OK. So that height from our green case plus the height from our blue case. OK. So once just looking at the distance from where the stone is dropped to the cliff and the other is looking from the cliff to the ground, we add those two and we get that total height. All right. So delta YC, we already know actually. OK. So we have that value check mark there. So what we need to find is delta YV. Now, in order to find delta YB, we're gonna need to know one more value A in green. We only have two known values V not and A. OK. But we have this relationship between the final velocity in B and the initial velocity in C. OK. So what we can do is calculate that speed at the cliff. Now we're gonna calculate the speed at the cliff which is at the intersection of these two, the speed at the cliff. And that's where we're gonna start. All righty. So let's do find the speed at the cliff. And we are gonna use our blue situation. OK. This post cliff section because we have three known values. We wanna find the Nazi. So we're gonna choose the kinematic equation, the Nazi A delta Y and T A, those three known values in the one we're looking for in that equation is going to be delta YC is equal to V, not C multiplied by TC plus one half, multiplied by ac multiplied by TC squared. Substituting in our values 30 m is equal to V Nazi multiplied by 1.58 seconds plus one half multiplied by 9.81 m per second squared, multiplied by 1.58 seconds off work. Let's go ahead and simplify. 30 m is equal to the Nazi multiplied by 1.58 seconds plus 12.2 4 4 8 4 2 m subtracting this 12.244842 m from both sides. And we get that V Nazi multiplied by 1.58 seconds is equal to 17.755 1 5 8 m. And finally dividing by that 1.58 seconds, we get that V Nazi is equal to 11.237 4 4 m per second. All right. So we not see that's the initial speed or velocity of the stone as it passes the cliff. OK. So the initial speed as it's going from the cliff downwards, remember is the final speed if we're looking at the stone from when it's dropped to when it hits the cliff, OK. It's this speed at the interface of those two motions. OK. So this is the speed we've just found, we can add it in to our information in green. OK. So this is our 11 0.237 or 4 m per second. And now we can calculate this delta YV value because we have VNVF and a three known values. OK. So we're gonna choose the kinematic equation with those four variables. We're gonna leave out the time because that's not what we're looking for and we don't have information about that. And so now we're gonna find the distance between the bird and the C. OK? All right. So the equation we're gonna use VFB squared is equal to V, not B squared plus two multiplied by A B multiplied by delta YB. We're solving for delta YB if we substitute in our values 11.23 7 4 4 m per second, all squared V knot is zero. OK. So this is gonna be equal to zero and then we have two multiplied by 9.81 m per second squared multiplied by delta YP. We can square on the left hand side and then divide everything by two, multiplied by 9.81. We get the Delta Libby, it's going to be equal to 6.4363 m. So let's go back to our diaper. We are looking for the total height from where the stone was dropped. We knew that the cliff had a height of 30 m and we found this speed at the interface which allows us to calculate the distance between where the stone was dropped to the cliff. Ok? So now all we need to do is add the two distances together the distance from the stone to the cliff and then the distance from the cliff to the ground is gonna give us that total distance or the total height that it was dropped from. All right, so let's do that. So now delta Y this total height is going to be equal to again, delta YB plus delta YC. This is gonna be 6.4363 m. What? 30 m, which gives us a height of 36.43 6 3 m comparing this to our answer choices. Ok? We're gonna round 23 significant digits. We can see that the correct answer is going to be option. A thanks everyone for watching. I hope this video helped you in the next one.

Struggling with Physics?

(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk.

(b) Assuming the balloons are being released from rest, from what height are they being released?

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Struggling with Physics?

(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk.(b) Assuming the balloons are being released from rest, from what height are they being released?

Watch next

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(II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk.

(b) Assuming the balloons are being released from rest, from what height are they being released?