You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance.

Video transcript

Hello everyone in this problem, you throw a stone vertically upward from the edge of a cliff whose height is 60 m. Stone reaches the starting .10 seconds after it was thrown, neglecting air resistance. What is the speed of the stone when it strikes the pond? Near the base of the cliff. Start by drawing a picture. We have some cliff and we also have a pond and the difference in height between us 60 m and we have some person all some initial velocity. It was just a maximum height H max B is equal to zero m/s and reaches back to its starting point After T is equal to 10 seconds. It's important to note that as the ball reaches back to its starting position, the velocity at this starting position is also the eye and we want to find V. F as it strikes the ground. We can split this problem into essentially two phases. First we want to find this V. I. As when we write the kinetic equations, the first one is V. F. Is equal to V. I plus 80. The second one is the F squared equals the I squared Plus two A Delta X. And the third one is delta X. Is equal to V I. T plus one, A half a T squared. We noticed that to find the F. We also always need a V. I. So this first phase can be used to solve for this V. I. Now in this first phase were given time and we know that it reaches its starting point in this full time. So the delta H actually would be zero as a final and age initial are the same. So we can use this third kinetic equation to solve for the eye. We have delta H, which is the same thing as delta X. We are looking for the eye, we have T. And we know that in free fall Acceleration due to gravity is negative 9.81 m/s square. We can rewrite this third cinematic equation as delta H is equal to the I. T. Plus one half G. T squared. And we know that delta H is just zero. So we can now solve for B. I. I. Is equal to negative 1/2 G times T. And we can make the substitution of these values negative 1/2 times negative 9.81 m per second squared times of tea that we're getting 10 seconds approximately gives us a value of meters per second. So now that we know B. I solve for V. F. In this second phase, In the second phase we have a V. I. And we have the we're looking for a V. F. And we have a delta H. We also know that it's in free fall, so G is equal still equal to make 9.81 m per second squared. So with these values we can use the second cinematic equation solve for the F. You can rewrite it as V F squared is equal to the I squared plus two G Delta H in the two. As this is different from the delta age. In the first version. Now we can simply solve from B. F by taking the square root of the right side Squared Plus two G. Other H. Two. And now we can make the substitution Where v. I was this 15 m/s. G is 9.81 m/s squared And H is 60 m. Actually this value we get that VF is 59.81 Years for a 2nd. His answer choice themselves have a great day.