00:01
All right, so in this problem, we're not told what the height of the cliff is.
00:07
Presumably, that's part of the example problem.
00:12
So i'm using 45 meters.
00:13
So anytime we plug in for the height, if it's not 45 meters, in your example problem, just substitute.
00:19
Now, the first part of this problem, we need to find the time.
00:24
And to do that, we need to get rid of the v -not in our vertical displacement equation.
00:29
We can do that by looking at the time part of the horizontal displacement.
00:35
And so the time that it takes to go from the cliff to distance d away can be found by taking d divided by the horizontal component to the velocity, b not cosine, beta.
00:46
Now we can look at the vertical displacement.
00:50
So it's going to move a distance of negative h from 0 to h.
00:55
And the vertical part of the velocity is going to be negative b .0, sine, theta, times this time.
01:03
And we're going to plug this in for time.
01:06
And then minus one -half gt squared.
01:08
It's minus because we're accelerating down.
01:12
Now when we plug this in for t, the b -nots cancel.
01:15
And we have sine over cosine, so we have tangent theta.
01:19
And so now we have this expression here, and we can then solve this for time.
01:25
And by solving this for time, we can just add this to both sides, add this to both sides, multiply through by two, and divide by g.
01:33
And we get this expression for time.
01:35
Oh, and then take the square root.
01:37
So we have two times negative d tangent theta plus h divided by g all under the square root.
01:43
Plug in our values for the distance, the angle, and the height, and we get a time of 1 .57 seconds.
01:56
It would be so slow.
01:58
There we go...