A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (c) When is the displacement of the boulder from its initial position zero?

Video transcript

Welcome back everybody. We have a rock that is thrown vertically upward which I will track its motion with this arrow and what goes up must come down. Now. We are told that its initial vertical velocity is equal to 98.1 m per second. And we know that at the very top of this motion it's going to come to a standstill for just a second. So we can say that our I know Vertical velocity is equal to 0m/s. Now, since this rock is in free fall as we are neglecting air resistance, the only acceleration acting on it is the acceleration due to gravity equal to negative 9.81 m per second squared. And we are tasked to find the time or the entire rip, meaning the time it takes to go up and the time it takes to go down. So first cinematic equation that pops into my mind for this problem is that our final vertical velocity is equal to our initial vertical velocity plus our vertical acceleration times time. We know some of these values. So let's go ahead and plug those in. Final vertical velocity we stated was zero. This is equal to our initial velocity of 98.1 Plus vertical acceleration of negative 9.81 times time attracting 98.1 on both sides. This yields that -98.1 is equal to negative 9.81. Now dividing by negative 9. both sides. We get that t equal to 10 seconds. Now hold on just a second is this the final T we're looking for? As a matter of fact, it's not this time is the time For only one direction, either up or down. So we are going to multiply both sides by two. To get that the T. Or time of our entire trip is equal to 20 seconds corresponding to our answer choice of the Thank you guys so much for watching. Hope this video helped, and we will see you all in the next one.