00:01
Hello, so for this problem, we are dealing with a baseball that's hit with an initial velocity, the knot, at an angle of beta knot above the ground, where our initial velocity is 140 feet per second, and the initial angle is 36 degrees.
00:30
Now before i move on and start solving the problem, i'm going to convert from feet to second to meters per second just so that i can then later on use the acceleration due to gravity g equal to 9 .8 meters per second squared and know that my units are always going to work out correctly because i'm using si units.
00:52
So 140 feet per second ends up being about 42 .672.
01:00
Meters per second.
01:03
Okay, now that we have that out of the way, we can jump right in for part a.
01:08
You're asked to find the x component of the initial velocity, v -0x, is going to be v -0 cosine of theta -0.
01:18
And when you plug in the values from the problem, you end up with about 35 meters.
01:24
I'm rounding all of my answers to two significant figures for the most part here, just so you know.
01:31
For part b, you're asked to find the y component of the initial velocity, so v -not -y will be equal to v -not times the sign of the angle theta, and when you plug in the values from the problem, you end up with an initial velocity in the y direction of about 25 meters.
01:54
For part c, you're asked to calculate the maximum height reached by the baseball.
02:04
And so to do this, i'm going to be using the kinematic equations for projectile motion.
02:12
And specifically, we aren't concerned with time and we don't know how long the ball is in the air before it reaches its maximum height.
02:20
So i'm going to use the kinematic equation that does not include time, which is that.
02:25
That v squared is equal to v not, or i guess vy squared is equal to v, not y squared, minus 2g delta y.
02:38
We're trying to find delta y, but the thing to note here is that when the ball is at its maximum height, the final velocity is going to be zero.
02:52
When it's at its maximum height, its velocity is zero.
02:55
So that means that 2g delta y is equal to v not y squared.
03:03
And we can solve for delta y here.
03:06
Delta y will be equal to v .0 y squared, which i'll go ahead and plug in our v .0 sine theta, not here, all divided by 2 times g.
03:22
And again, i used 9 .8 meters per second square.
03:25
For g when you plug in the values from the problem you end up with a maximum height reached of about of about 32 meters for part d here you're asked to calculate the time it takes for the baseball to reach its maximum height so now we're going to want to use one of those other kinematic equations because we're solving for time i'm going to use the kinematic equation where you don't have time squared in it because it'll be a little bit easier to solve.
04:02
And so that is v is equal to v not minus g t.
04:08
And again, i should add y's here.
04:10
So vy is equal to v.
04:12
Not y minus gt.
04:14
And we're trying to solve for time in this case.
04:19
But again, our final velocity is going to be zero in the y direction when we're at our maximum height.
04:27
And so you can maybe see here that t is going to be equal to the initial velocity in the y direction, which is v -0 -sign -theta -not divided by g.
04:42
And when you plug in the values from the problem, you end up with 2 .6 seconds.
04:49
Roundabouts for that.
04:52
For part e, you're asked to calculate the height reached by the baseball in one second.
04:58
So we are going to finally use that last kinematic equation in order to figure out that height.
05:08
So delta y is equal to v .0y t minus one half gt squared.
05:20
And we're going to plug in all of the values from the problem...