00:02
Welcome to this numerate tutorial.
00:04
So we have our airplane and it drops a projectile at a 53 degree angle to the horizontal.
00:18
And then there's a descent of 730 meters.
00:22
So we can use the delta y is equal to initial velocity times time plus one half gt squared formula to describe the initial velocity prior to release plus the one -half gt squared descent for the v -sub -y velocity on the y -axis vertical so there's our projectile deployed at the specified altitude 730 meters above ground level and then we have our downward vertical vector representing gravity, and then we have her vx at specified angle 53 degrees.
01:12
So if we solve for v sub o, we simply can find its value algebraically by solving for v sub o.
01:26
And upon doing so, we compute 121 .50 meters per second.
01:35
So that would be the answer for part a of the problem.
01:43
So now we continue with its flight path and note that because it's not deployed at zero degrees equal to the horizontal, there will be a parabolic travel as the projectile then rotates downward to 90 degrees and makes impact with the ground.
02:05
So we must get an idea of just the vertical component, the vertical velocity component, so that is equal to acceleration.
02:19
In this case, gravitational acceleration times time.
02:23
We have a five second impact time, so 9 .81 meters per second squared, gravitational acceleration times five seconds.
02:32
So our velocity at impact is 49 .0 meters per second.
02:40
And that is just a reference to realize the vertical component aspect.
02:50
And remember that the projectile will remain its velocity at deployment.
02:57
Initial velocity will remain constant.
03:00
So v .o is equal as v.
03:03
Sub f.
03:04
Okay, so if we draw a graphic to compare the arc of descent relationship between vertical and horizontal components at zero degrees versus 53 degrees, we get a graphic similar to this rendition...