00:01
So the question states that a projectile is launched at 15 degrees, and the highest barrier it can go over is 13 .5 meters high.
00:12
And so we need to try and find what this velocity v is equal to so that it goes over at its maximum point.
00:20
So its maximum point is 13 .5 meters high.
00:25
So to do this, we first can separate our velocity into two components.
00:30
In this case, we just need to focus on the vertical component of the velocity, which we can find because we know sign of 15 degrees is equal to the opposite over hypotenuse.
00:45
So the opposite in this case is the vertical component of the velocity over the hypotenuse, which is just the velocity, which means that the vertical component is equal to v times sine of 15 degrees.
01:01
And now that we know this, we can use our kinematics equation, which states that the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the displacement.
01:18
So we know when the projectile is at its maximum, its vertical velocity will be zero.
01:27
We know what the initial vertical velocity is of the projectile, which is v -sign.
01:34
15 degrees.
01:36
This will be squared.
01:40
And we know that the acceleration is going to be negative 9 .8 meters per second squared due to gravity...
