A rubber ball is shot straight up from the ground with speed v₀. ...

Question

A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.a. At what height above the ground do the balls collide? Your answer will be an algebraic expression in terms of h, v₀, and g.

Accepted Answer

Hey, everyone in this problem. A box is released from rest from a balcony at the same instant, a worker standing below the balcony throws a ball vertically upwards with an initial speed of 8 m per second. The distance separating the hands of the worker in the balcony is 10 m. Ok? So we're gonna come back to this problem and figure out what it's asking us. But first we're gonna draw out this situation, it's given us, we have the balcony and we're gonna have a box that is released from that balcony, ok? And because it's released, the initial velocity of the box is gonna be zero. OK? So we're gonna use subscripts one to denote the box. So V not one is gonna be 0 m per se and it's gonna be released downwards. Then we have the ball. Yeah, it's gonna be thrown by a worker and it's gonna be thrown upwards vertically. And the initial velocity or the initial speed we're told is 8 m per second. Ok? So we're gonna take upwards to be a positive direction. So that's gonna be a positive initial velocity. This is gonna be V not two. So that subscript two indicating the ball of 8 m per second. Now, the other thing we're told here is that the distance separating where the box is dropping from the balcony in the hands of the worker is m. Ok. So the initial distance between the two is 10 m. All right. Now, getting back to this question, we're asked to determine the position with the respect to the hand of the worker where the ball and the box meet. Then we have four answer choices all in meters. Option A 2.34 option B 4. option C 6.67 and option D 7.21. Now this is a kinematics type problem. OK. We have this box following the ball being thrown. So let's think about the variables we have for both the box and for the ball. So for the box, we've already said that the initial velocity of V not one is 0 m per second. We don't know what the final velocity will be when it hits the ball. We know that the acceleration is gonna be the acceleration due to gravity. It's acting downwards. We've chosen up to be a positive direction. So this is gonna be a negative 9.8 m per second squared. Don't know why one, we're gonna leave for now and T one we're gonna leave for now as one. And we don't know the actual numerical values of these, but we're gonna come back to that in just a second over the ball. Same thing V 02. That initial velocity is 8 m per second. We don't know the final velocity of the F two. We're gonna have the same acceleration due to gravity acting on the ball as the box. So we have negative 9.8 m per second squared for a two. We don't know delta Y two and we don't know T two. Now, if we look at the information, we have all we know is two things from each object. We have V not an A for the box and for the ball and that is not enough to solve for any of these other variants. Mhm. However, we can relate the two. OK? If we consider that delta Y two is gonna be the position with respect to the sorry, with respect to the hand of the worker where the ball involves me. OK? Then that's gonna be the value we're looking. So we're looking for delta Y two and that's gonna be the distance from the ball upwards to where it contacts the ball. Now, if that's the case, we can write delta Y one in terms of delta Y two. With the information, we know about the initial distance between the box and the ball. It is 10 m. So we know that the displacement for the box is gonna be negative because the box is falling downwards and upwards is our positive direction and the distance, well, the distance of travels is gonna be that total 10 m minus whatever this delta Y to it, say the distance from the ball's initial point up to where the ball unbox meet. So now we've written delta Y one in terms of delta Y two and we can do the same with T one. OK. The time when they meet is gonna be the same whether we're talking about the box or the ball. This time point is the same, we know that T one is equal to T two. Now, what we can do is write two equations with 21 notes. OK. So we have this delta Y two value and this T two value that we don't know, but it's consistent in both of these equations. We can go ahead and do that. So starting with the box, right? And recall, we're gonna choose the U AM equation that does not include the final velocity. We have no information about that at all. So we're gonna leave that out and use the other variables that we either know or have some information about the relationship. OK? So we have delta Y one is equal to V multiplied by T one plus one half a one multiplied by T one squared substituting in our values, we have negative 10 m minus delta Y two is equal to and the initial velocity is zero. So the first term goes to zero. We're just left with one half multiplied by negative 9. m per second squared multiplied by T two square. Now, we have two variables, delta Y two and T two. We're gonna use the substitution method. So what we wanna do is write this equation in terms of one of those variables. It's gonna be a little bit easier to do delta Y two. We don't have to worry about taking a square root or anything like that. So let's go ahead and isolate this equation for delta Y two. First thing we can do is multiply both sides by a negative. So we have 10 m minus delta Y two is equal to 4.9 meters per second squared. OK. That negative goes away because we're multiplying by another negative multiplied by T two square. All right. Now we're gonna move this 10 m to the other side that gives us negative delta Y two is equal to 4.9 m per second squared, multiplied by T two squared minus 10 m. And finally dividing by this negative, we get delta Y two is equal to negative 4.9 m per second squared, multiplied by T two squared less 10 m. And we're gonna call this equation what? OK. And this is the first equation we're gonna use in our substitution method. And the second equation is gonna come from the ball. So let's go back to the ball and we have the same information. OK. So we're gonna use the same kinematic equation. We're gonna ignore the final velocity and choose the equation with it because we don't have information about that. And that equation is going to be delta Y two is equal to V, not two multiplied by T two. That's one half a two, T two squared substituting in our values, we get delta Y two is equal to 8 m per second multiplied by T two. That one half multiplied by negative 9. m per second squared multiplied by T two squared. And this is gonna be equation two. OK. So now we have our two equations with two unknowns, delta Y two and T two. We're gonna substitute the first equation into the second equation and solve for one of these variables. So substituting equation one into equation two. What do we have? OK. Well, equation one is delta Y two is equal to some stuff. Equation two is delta Y two is equal to some stuff. And so what happens is the stuff is gonna end up being equal and we have negative 4.9 m per second squared, multiplied by T two squared or less 10 m going to be equal to 8 m per second, multiplied by T two minus 4.9 m per second squared multiplied by T two squared. So we have this negative 4.9 m per second squared, T two squared term on each side and so we can move either one to the other side and they are gonna cancel each other and that term is gonna act as zero. It's gonna go away. And what we're gonna be left with is that 10 m is equal to m per second, multiplied by T two, dividing both sides by 8 m per second. We get that T two is equal to 1.257. All right. Now, we have to think about what this time is. OK? We're not looking for the time that they meet. We're looking for the distance, OK? Or the position with respect to the worker. OK? Now, we figured out this T two value which is the time where the ball and the box meet. So now that we know the time we can go ahead, substitute that back into either equation one or two to find the corresponding position. Now I'm gonna choose to substitute that back into equation one because equation one is just a little bit simpler, but you could substitute it into either equation and you're gonna get the same result. So we're gonna substitute T equals one 0.25 seconds into equation one. And when we do that, we get the delta Y two is equal to negative 4.9 m per second squared, multiplied by T two squared plus 10 m negative 4.9 m per second squared multiplied by 1.25 seconds squared. Plus 10 m. And if we work this out, we get the delta Y two is going to be equal to 2. m. OK? So that is the position we were looking for if we go back to our diagram and remind ourselves of that position and that position is the position or the displacement change in position of the ball. OK? As it gets thrown upwards, which is gonna be the exact same as the position with respect to the hand of the worker where they meet and that box and that ball. So comparing this to our answer choices, we can see rounded to two decimal places. We have answered choice. A 2.34 m. Thanks everyone for watching. I hope this video helped you in the next one.

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A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.a. At what height above the ground do the balls collide? Your answer will be an algebraic expression in terms of h, v₀, and g.

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