0:00
Hi there.
00:01
So for this problem, let's start with problem five.
00:03
Proble five says that a football kick in the air from ground level reaches a height of 25 meters and lands 83 meters away.
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What was the initial velocity? so we are told that the maximum height for this was 25 and that the range for this is the water that adds is just 83 meters away.
00:32
At the initial height for this, it was zero since it is from the ground level.
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And we are asked from this to determine the initial velocity.
00:47
So the initial velocity has two components, okay? so we need to find those.
00:52
For this, we can use first the expression that adds the horizontal range is the initial speed times the cosine of d -tot this times the time.
01:01
And the vertical motion is just that y is equal to the initial speed the sign of theta times a time minus 1 divided 2, the acceleration to do gravity times a time square.
01:13
So once we have this, what we can do is we already have ads, but we don't know the value of the cosine of theta or the angle theta, or the initial speed.
01:32
Hi there.
01:32
So these two equations are not the ones that we are going to use.
01:35
We're going to use an equation from kinematics, and in this we are going to use the vertical, component of the velocity.
01:43
That will be the initial vertical component of the velocity to the square minus two times acceleration due to property times we're going to use the situation when we are at the maximum height.
01:53
Why we use this? well, because we know that the final speed at that moment is just equal to zero.
01:59
So in here we can solve for the initial speed in the vertical component.
02:05
So there will be two times acceleration times the maximum height and we take the square root of that.
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So then we just substitute the values, the square root of two times 9 .81, the height of this, which is 25, and which is square root of this.
02:20
So then using our calculator, we obtain a value of.
02:28
So the value that we obtain for this is 22 .15 meters per second.
02:39
Okay.
02:41
So that is the vertical company.
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Now, to obtain the horizontal company we can use, we can obtain first.
02:48
The time to reach that maximum height.
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So that will be the initial speed divided by the acceleration due to gravity.
02:53
So that will be the 22 .15 value that we just obtained divided by 9 .81.
02:58
This will give us a value of about 2 .26 seconds.
03:01
And then we know that the total time is the double, doubles this time, okay? because it's twice the time up.
03:08
This is the time up.
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So that will be two times the value that we just obtained.
03:14
That will give us 5 .4 .52 meters.
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Seconds.
03:22
And once we have this, we need with that, we obtain that the horizontal, well, the reach, the horizontal reach is just the edge of the speed times the time.
03:33
So we will obtain the x component of this.
03:35
That will be x divided by the time.
03:37
So that will be the value that we are given, that is 83.
03:40
And then this divided by 4 .52.
03:44
So that will give us a value of 18 .37 meters per second.
03:48
So once we have that, since what we won in this case is what was the initial velocity.
03:56
The initial velocity in a vector form is the x component that is 18 .37 in the x component plus 22 .15 in the y component.
04:11
So that will be the solution for problem 5.
04:16
And upper problem sets.
04:19
We are asked about where the cannonbell is launched in the air from ground level...