00:01
Here we're looking for a special angle where the range of a projectile that is launched from the ground, returning back to the ground, that that range distance in the x direction is equal to the maximum height that the projectile achieves at half the time.
00:25
So is there a theta knot or find the theta knot? such that r equals y max.
00:40
So that's kind of what the picture is trying to show that those two distances are equal.
00:47
Now we do have an expression that does not involve time for the range of a projectile.
00:54
It involves both the x and the y components of the velocity.
01:00
So the sine of theta zero and the cosine of and i won't derive that.
01:12
It is not too hard to derive, but we have to remove the time from our kinematics formulas.
01:20
Likewise, we can find an expression for ymax, again, using kinematics, which we won't go through, but it just involves the square of the y component of velocity, and then over 2g.
01:44
And so we're simply going to set those equal to each other and solve for the special theta -0.
02:01
And when we do that, a lot of things will cancel.
02:05
We want to be a little bit careful with that, but we have 2 v -0 squared for the range.
02:11
Sine of theta -0, cosine of theta -0, over g is equal to v -0 squared.
02:23
Sine squared of theta 0 over 2g.
02:31
So a lot of things will cancel.
02:34
One factor of sine goes away.
02:38
The v0 squared goes away, and notice that the g goes away.
02:44
And if i put all the numbers together on one side, we have four, which is two times two, is equal to sine of theta not over cosine of theta -not.
03:00
And that is just the tangent of this special angle.
03:06
So we can definitely solve for this special angle by taking the inverse tangent of four.
03:18
Right.
03:19
So just invert a function to get rid of the function on one side and put it on the other.
03:26
And this basically comes out to be 76 degrees.
03:30
We'll call it 76 .0 degrees to the nearest precision...