00:01
In this problem, we are building a tributchet, which is a catapult -like weapon, and we want to launch stones off a 75 -meter high cliff with an initial speed of 36 meters per second.
00:11
Some people say that a 45 -degree launch angle would give the maximum range, but other people are saying that the claim, that the cliff height would make a difference.
00:19
So we want to go ahead and figure out what angle would maximize the range.
00:25
So we have a cliff, and we have our castle down here.
00:29
So we're going to launch, and we want it to hit like a little.
00:32
This.
00:35
So to do that, we have, we have our angle.
00:43
So the x and y coordinates of the projectile with an initial height of h and an initial velocity of v, we x is equal to v cosine theta times time, y is one half g, t squared, plus b sine of theta, times t plus the height.
01:06
And those are just from the kinematic equations.
01:10
So one method is to use the expression for y to find an expression for the time when the projectile lands on the ground.
01:16
Then use this time to find an expression for the horizontal position of the projectile when it hits the ground.
01:22
And we'll find the maximum value for this range expression by taking the derivative, and you found the maximum range in an angle that gives the maximum range.
01:31
This way leads to a lot of algebra, so we'll use a different way.
01:36
So in general, for a given initial speed, there are two initial angles that will cause the projectile to land in the same place.
01:41
However, at the maximum range, there's only one initial angle.
01:45
So we'll use this after setting up the algebra.
01:47
So we're going to solve the x equation for t and substitute that into the y equation, which will give us this...