Nicole throws a ball straight up. Chad watches the ball from a window 5.0 m above the point where Nicole released it. The ball passes Chad on the way up, and it has a speed of 10 m/s as it passes him on the way back down. How fast did Nicole throw the ball?

Video transcript

Hi everyone in this practice problem, we will have an athlete tossing a ball vertically upward with a speed of P one. A velocity sensor is placed exactly at three m above the point of release of the ball measuring a speed of eight m per second. When the ball is at the same horizontal level as the sensor moving upward, we're being asked to calculate what the initial velocity fee one is that the athlete is applying to the ball. The answer choices are a 2.3 m per second. B 7.7 m per seconds, C eight m per second and D 11 m per second. So first, I am going to draw our system which we have. So we have a point of initial point which is Y one equals zero m and V one, which is the one that's being asked. This is essentially is the release point of the ball. So I'm gonna call that the release point. And then essentially we have the ball with the velocity sensor here and that will be at a point of three m above to a release point where a measure velocity feed 2 of 8 m/s. And we wanna recall that um there will also be a gravitational acceleration applied to this motion. So we will model the ball as a particle under constant acceleration, which is going to be the gravitational acceleration. Here. We will use the kinematic equation to calculate for our initial speed P I or pe one which is going to be P two squared minus P one squared equals minus two G Y two minus Y one. This is going to be the kinematic equation that we're gonna use in order for us to solve for this problem. So we wanna rearrange this so that we can find what P one is. So P one is then going to equals to P two squared plus two G Y two minus Y one. So next, we wanna actually input all the unknown information. So P one squared will the equals to B P two squared which is eight m per second squared plus two G which is two multiplied by 9.81 m per second squared. And Y two is going to be three m and Y one is going to be the release point which is zero m. This will give us a V one squared value of 122.86 m squared per second squared. Then because of this V one is then going to be the square root of 122 oops .86 m. Squared per second squared, which will equals to approximately around 11 m/s. So the athlete essentially tossed the ball vertically upwards with a V one speed of 11 m per second, which will equals or correspond to option D in our answer choices. So answer D with the initial P one vertical speed of 11 m per second is going to be the answer to this problem statement. If you guys still have any sort of confusion, please make sure to check out our eyeglass and videos on similar topics and that will be all for this one. Thank you.