00:01
A grasshopper on a horizontal plane jumps with an initial speed of 3 .40 majors per second in a direction of 43 degrees above the horizontal.
00:09
At what distance will it land? so we have the grasshopper at, we'll say, this point here, and then it, we know it travels in a direction of 43 degrees above the horizontal, so we'll let this be the horizontal.
00:26
So there's our 43 degree angle.
00:28
So we are going to split that vector, the 3 .40, into its horizontal and vertical components.
00:40
So that's a velocity vector.
00:42
It has a horizontal component and a vertical component.
00:46
We'll call this one the vertical vy and the horizontal vx.
00:51
So if we do some trigonometry, we can see that the sine of the 43 degree angle is vy over the hypotenuse of 3 .40, and then the cosine of the 43 degree angle is vx over 3 .40.
01:14
If we solve for vy, that would be vy, we multiply both sides by 3 .40, that gives us 3 .40 times the sine of 43 degrees, and then vx, we multiply both sides by 3 .40, that gives us vx to be 3 .40 times the cosine of 43 degrees.
01:35
So we're going to need those components of the velocity right there, the vertical vy and the horizontal vx.
01:43
So now, we need to know what distance will this land since the grasshopper will go up and then eventually it's going to come back down.
01:53
So we need to find at what time will its height be zero, and also we're going to need the horizontal motion equation.
02:02
So the horizontal component, the vx, that's not going to change.
02:08
The acceleration due to gravity is only affecting the vy.
02:11
So our horizontal velocity will just be a constant, so the horizontal distance, we can treat that like just a movement with a constant velocity, that would be the velocity times time.
02:28
So the horizontal velocity times time will tell us what the x or horizontal distance is.
02:36
So we just want to know what is that x value there, how far does it travel horizontally, and we need to find the time that it hits the ground...