00:01
All right, so let's say a cannonball is fired at a 50 degree angle, and it's fired from an initial height of one meter.
00:11
And let's see, the cannonball travels, we're told, a horizontal distance, call this x of 60 meters, and then passes over a castle wall by two meters.
00:23
So the height at this point is two meters.
00:26
And so we wonder how long does it take for the cannonball to reach the wall? all right, and actually, sorry, it's two meters above the wall, which is 10 meters tall, so we'll write this as 12 meters.
00:36
So we can use the static kinematic equation to do this, which says that the displacement of the object in the y direction is going to be equal to x times the tangent of the launch angle, minus one half g x over v squared times one plus the tangent squared of theta.
00:57
And we don't know v, but we know everything else, so we can solve for the velocity.
01:00
And delta y if you look at it it's the you know it's the displacement on the y direction so it'll be 11 meters and so if we kind of rearrange this we'll have one half g x over v squared times one plus the tangent squared of theta equals and we'll write it as delta h like the change in higher i guess delta y will keep it consistent delta y uh minus the uh tangent or minus x times the tangent of and so if we solve this equation for v, what we should have, and i'm missing in parentheses, is g x squared times 1 plus the tangent squared of theta over 2 times delta y minus x times the tangent of theta...