00:01
So in this question, we have a batter who hits a ball directly upwards at home plate with some speed v, and the ball is caught six seconds after it struck.
00:13
So delta t is 6 .0 seconds.
00:15
That's how long it's in the air for.
00:18
Well, let's think about the, so the x velocity of this is always going to be zero.
00:22
There's no force in the x direction.
00:25
But the velocity in the y direction is going to be v minus gt.
00:31
And that's because it's accelerating due to gravity in the y direction.
00:36
So what we can do is we can integrate this expression to get y of t is vt minus a half gt squared.
00:46
And what we know is that y of zero equals zero.
00:50
We can see that right here.
00:52
But we know that y of delta t is also equal to zero because that's when it has come back down and landed in the position where it was to begin with.
01:00
So this is v delta t minus a half g delta t squared.
01:06
So that tells us that v, sorry, that's not a v, v is a half g delta t squared divided by delta t.
01:17
So i've moved this over to the other side of the equation, then i divide by delta t, and that means i can cancel out that square.
01:25
So that tells us that the initial velocity is 0 .5 times 9 .81 times six seconds.
01:34
Which is 29 .43 meters per second.
01:40
So that's the initial velocity of the ball.
01:43
And now we also know that y of t is, well, we've got it written down up here, but now we know what v is.
01:51
So we know everything about this ball's trajectory now.
01:56
So part b, what's the maximum height the ball reaches? so when the ball's at its maximum height, its velocity is equal to zero.
02:07
So v of t, max equals 0, but we know that the velocity in terms of time is v minus g t...