A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake?

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Hey everyone in this problem, a diver awaiting his turn to jump from a bridge, awkwardly drops his phone. Initially, the mobile phone is located seven m above the river surface. At the river surface. The mobile phone has an impact speed of V. We're going to assume that the mobile phone continues its motion vertically downwards with a steady speed of V until it reaches the ground at the bottom of the river. The time taken to reach the bottom of the river, measured from the instant the phone was dropped is six seconds and were asked to determine the river's water level. Alright, so let's draw a little picture. So we have the bridge up here and we have a river down below and the the water level of this river we're gonna say is called R and that's what we're looking for. We're looking for the water level of the river and we know that the distance from the bridge to the top of the water is seven m. So let's take down to be the positive direction. And let's write out what we know And we're gonna consider this a two part problem. So let's first look at what happens when we drop the phone until it hits the top of the river. Okay. The surface of the water. So the initial speed be not we're gonna call it V not one. This is gonna be 0m/s. Okay, we drop it. So initially it has no initial speed. The final speed. V. F. One we're told is vi okay, it has an impact speed of V but we don't know what V is. The acceleration is going to be the acceleration due to gravity. And we've chosen down to be the positive Y direction. And so the acceleration is going to be positive 9.8 m per second squared. We don't know how much time this takes. And the vertical distance, the change in vertical distance is going to be seven m. Any of the phone falls seven m to the top of that reverse surface that's downwards are positive direction. And so delta Y one is seven m. So we know three pieces of information. We don't know VF one and we don't know t one which of those do we want to find out once this the phone is in the river, we're told that it has a steady speed. That means the acceleration is zero. And so we can consider in that phase that the velocity is equal to the distance over time. Okay, When the acceleration is equal to zero. So if we consider this then in order to determine the water level, which is gonna be the distance it travels. In that phase, we need to know the speed that it's traveling through the water and the amount of time it takes. Okay, so we actually need to know both pieces of information. So let's start by finding the speed. V. So VF one square. Okay, we're going to choose the U. A. M. Equation. That doesn't include T VF one squared is equal to v. Not one squared Plus two. A 1 delta Y. One VF one. Well that's just v squared. A VF one squared is B squared. The initial speed is zero. So that term goes to zero, we get two times 9.8 m per second squared times seven m. This gives us V squared is equal to 137.2 m squared per second squared. And when we take the square root we're going to get a positive and a negative answer. We know that the velocity is going downwards downwards as a positive direction and so the velocity is going to be positive. So we're gonna take the positive route and that is going to be 11.713- m/s. Alright, so we know the speed and that impact speed V is the same speed that it carries through the entire river. So we know our speed V. Now let's figure out the time it takes for the phone to get to the river surface in the problem, we're told the total time from when it's dropped to it hits the bottom of the river. Okay, So if we can figure out how much time it takes to get to the river surface then we'll know how much time it spends in the water itself. Okay. And so we're gonna choose the um equation here without VF. Okay, we know V not a in delta Y. Now we've also just calculated VF but it is a bit of an approximation we've had to round um and cut off some digits. So it's always best if you have that situation to try to use the ones that you were given in the problem instead of the ones that might have a teeny bit of rounding. Okay. And so we're gonna choose delta Y. One is equal to V. Not one T one plus one half a one T one squared. So we have that seven m is equal to 1/ Times 9.8 m/s squared times T one squared. Okay, this v nought T one term went away because V not one is zero. If we isolate for T one squared case we divide by one half times 9.8 we get that T one squared is equal to 1.42857. In our unit meter divided by meter per second squared. So we get second squared and when we take the square root again we're gonna get the positive and negative root. But when we're talking about time It's a positive value. And so we're gonna take the positive route and it's gonna be 1.19523 seconds. So this is the time that it takes for the phone to fall from the bridge to the top of the water surface. And so when we go into this constant speed phase again we can use the velocity equals distance over time equation. We know that our speed or velocity is 11.713-4 m/s. Okay, that's the impact speed on the river, which is the speed it carries throughout the entire river. We're told that in the problem we want to know the river. Okay. The distance that it travels is a distance through the river. And so that's the value are that we're looking for And the time that we spend in the water. Well, the total time from the top of the bridge to the bottom of the river is six seconds And 1.19523 seconds of that are spent getting to the water surface. Okay, so we have six seconds minus that time it takes to get to the surface is gonna tell us how much time the phone is actually in the water. And when we multiply we get a distance r of 56.28 m. Okay? And so the water level of that river is gonna be 56. m. If we go back to our answer choices. Okay, We round to the nearest meter, we see that that's going to correspond with answer choice C m. Thanks everyone for watching. I hope this video helped see you in the next one

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake?

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Key Concepts

Free Fall and Gravitational Acceleration

Constant Velocity in Fluids

Total Distance Calculation