00:01
So in this problem, we have a baseball being thrown off of a cliff.
00:07
And so the baseball goes some distance and then hits the ground.
00:12
So what we are told is that, firstly, the baseball is thrown two meters off the top of the cliff.
00:21
And so we're going to say this cliff has some height h, which is what we want to find in the end.
00:26
And then if we continue the two meters up further, the ball crosses the point that it was launched at or thrown at.
00:35
At 0 .51 seconds after it was thrown.
00:41
We know that the initial velocity of the ball at some angle, we don't know the angle, but the initial velocity was 34 .5 meters per second, and the final distance that it traveled was 125 meters.
00:58
And so what we need to do is break this problem into a few parts, where we first figure out some information about the initial velocity based on our data point here and then we can use that to find the final or the final height based on the total flight.
01:17
So what we're going to do is set an axis here where the origin is placed at the place of the ball being launched.
01:28
And our first part we consider this top section where the ball is thrown and returns back to its launch altitude.
01:38
So in this case, yep, so in this case, we can imagine that the ball, instead of traveling to the side, went straight up and straight down.
01:49
And if it did so with the same initial velocity in the y, when the ball landed, it would be going downwards with the same velocities, so negative v0 .y.
02:00
And we know that this took 0 .51 seconds, and the acceleration of gravity here, g, whereas on earth.
02:07
And so we can use the kinematic equation.
02:10
V is equal to v0 plus a t.
02:15
And as we talked about in our little picture here, we have our final velocity being negative v -0.
02:22
Our initial velocity, which we want to find, these are both v -0 -y.
02:26
The acceleration of gravity here is negative g because it's pointed downwards relative to axis.
02:32
We're talking with our red axis from above this one, so y is upward, so g pointed downwards is negative.
02:39
Then what we can do is solve this for v -not -y.
02:43
And when we do that, we end up with this, we can start plugging in what we have over negative 2.
02:58
And so our initial velocity in the y component turns out to be 2 .50 meters per second.
03:06
Now in this problem, there's a few steps to go, so it's really helpful to care through this number in a calculator.
03:12
So you get a little bit more position.
03:14
If not, maybe add a couple more digits to make sure your answer stays correct.
03:20
So now what we know is that the ball was launched at some angle, 34 .5 meters per second, that its y component was 2 .5 meters per second.
03:32
So what was its x component, v .0 .x? well, we see we have three of the sides of a, two of the sides of a triangle.
03:40
So we can use pythagorean theorem to find the other one.
03:44
So initial velocity in the x would be equal to when we solve pythagorean theorem, 34 .5 squared is equal to 2 .5 squared plus v.
03:52
V .0x squared, we'll see that v .0x, actually, is equal to 34 .5 squared minus 2 .5 squared...