A rookie quarterback throws a football with an initial upward velocity component of 12.0 m/s and a horizontal velocity component of 20.0 m/s. Ignore air resistance. (b) How high is this point?

Video transcript

Welcome back everybody. We are told that a soft ball is thrown with an initial horizontal velocity of 15 m/s. And we are also told that it's vertical Initial velocity is equal to that. So the Y component is also m/s. That means we are looking at a velocity vector, an initial velocity vector of something like this. Now this softball is going to follow projectile motion means it's going to follow an arc that's parabolic kind of like that. And at some point up here it is going to reach its maximum and we need to figure out the distance from the ground designated by Y max at that maximum point. Well, our why max here is going to be equal to one half times our acceleration due to gravity times our time. We know this, that's just an equation were given because we're dealing with projectile motion. But what is our time Here? Well, our time here is our time it takes to reach that maximum height. Now this is given by another equation which is equal to our initial vertical velocity over our acceleration due to gravity. So now that we have these equations, let's go ahead and plug some values in our T max is equal to our initial vertical velocity of 15 divided by our acceleration due to gravity. And this is going to give us 1.53 seconds to reach its maximum height. Now let's go ahead and plug that into our maximum height formula. We have one half times the acceleration due to gravity, times our time that we just found squared. And when you plug this into your calculator, we get 11.48 m corresponding to our answer choice of a thank you guys so much for watching. Hope this video helped, and we will see you all in the next one.