![]() ![]() Times the sine of 53 degrees, times our change in time, The first time we doĪ new type of problem, it's good to know- so 90 And remember, it's negativeīecause our displacement is going to be downwards. Write the units here just so that we don't take up Negative 16 meters, right? Because 25 minus 9 is 16. The vertical direction is going to be equal This thing is traveling, it will be displaced Out how long were in the air? So what is the displacement? If we're starting In time plus the acceleration, times the change in time The vertical direction right here- times the change We know that the displacement isĮqual to the initial velocity- and we're dealing with Let me just copy and paste this a little bit We can use the formula that we derived in Or the same magnitudeĪnd velocity but the opposite direction. Off at, it's going to be the same velocity but ![]() Reasoning, that hey, whatever velocity we start And especially since we'reĭealing with different levels, we can't use that more basic Horizontal component is equal to 90 timesįigure out how long this thing stays in the air? Well, we'll use the Over the hypotenuse, over 90, is equal to the Cosine is adjacentĬomponent of our velocity, I'll say in the x direction, The horizontal component, the horizontal side Magnitude of that side is going to be equal to 90 Over the magnitude of our original vector. That's the verticalĭirection, over the length of the hypotenuse, It's the magnitude of the vertical velocity. Opposite side, is equal to the vertical velocity. ![]() Trigonometry sine of an angle is opposite This side right over here? Well, this is the opposite side. The vertical componentĬomponent of the vector- what would be the length of We're going to assume that air resistanceĭid in the last video- and I'll go through all of the And then you use the horizontalĬomponent to figure out, given how long it's in theĪir, how far did it travel. You use the verticalĬomponent to figure out how long it's going Vector into its horizontal and vertical components. You'll always want to do is divide your velocity Vertical direction as soon as it leaves the muzzle. Meters, because that's when it's leaving the muzzle. We're launching it from an altitude of 25 I know in the last video,Įven though I drew the cannon like this, we Launching this from a height of 25 meters. Over here, let's say that that is 25 meters. Launched from- so from the muzzle of theĬannon down to here. And just to give ourselvesĪ sense of the heights, or how high it's being To have it come out of the muzzle of the cannon Let me do it this way, just to make it 100% clear. Let me draw this aįire the projectile at an angle of- let me To launch the projectile off of a platform. ![]() More complicated two-dimensional projectile Vf^2=Vi^2+2ad (Vi=90sin53 and a=9.8 and d=16) so Vf=74.03m/s.Now its is time to find how long it took to fall this 16m.then Vf=Vi+at (Vf=74.03 Vi=71.88 a=9.8)and a=+9.8 because gravity is helping its Velocity =0.22s (time it takes to fall the 16m).final time in the air will be 14.67s+0.22= 14.89s.then to get the X-distance just use D=V*t.(D=90cos53*14.89= 806.5m.Sorry for the long winded explanation. There is another way (using basic algebra.takes forever).For Y(direction).You will need to use.Vf=Vi+at (Vf=0 Vi=90sin53 a=-9.8) t=7.33(this is just the time up) but total air time would be (7.3344*2= 14.67s).to land on its original horizontal axis.and a=-9.8 because the object is going up and gravity is slowing its upward velocity.to where Vf=0 at 7.3344s).Now as the objects falls, since the y-direction s as if Vi=0 and Vf=90sin53.as it returns to its horizontal reference.BUT once it falls below it.we have a new calculation.to consider.Our Vi=90sin53 because it has an initial velocity but we now do not know its new final velocity.We have to find it. He knows that Juniors and Seniors in high school and college know the quadratic equation. Sal, worked this problem in such a way to avoid using several other algebraic equations. ![]()
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