00:01
So we're given these parametric curves for the equations of a projectile and we're told in part a that v0 is 500, alpha is 30 degrees, and g is 9 .8 meters per second squared.
00:27
So first of all we want to find the time.
00:32
To get the time we take the y equation, we put in y equals 0.
00:40
So it's v0 sine alpha minus half g t times t.
00:51
I'm just factoring out a t from that.
00:59
So that's a quadratic.
01:02
It has two solutions.
01:04
The first one is t equals 0.
01:07
That corresponds to our starting point.
01:09
This other solution, which is the one we want, is v0 sine alpha over half g.
01:28
So sine of alpha, alpha is 30 degrees, is the square root of 3 over 2.
01:45
Actually it's a half...