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For number 42, we're given a projectile is launched at 30 meters per second and angle 60, and it's in the air for 7 .5 seconds.
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And we're asked to find, first of all, how far above or below the launch point.
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It lands.
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So i know i'm just working with the vertical part here, so i'm working with the y.
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I know my initial velocity will be this component.
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It.
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So that's the opposite side.
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So i'm going to do sign of 60 times the hypotenuse, which is 30.
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And i get that initial velocity is 26 meters per second.
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I know acceleration is negative g, and i'm looking for the displacement.
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So i'm going to use my displacement equation.
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And i get negative 80 .6.
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So the negative just means it's below the starting point.
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So maybe it went up and it came down to here somewhere.
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So i may ask a distance, not a displacement.
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So my answer with just two sig figs will be 81 meters.
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Part b, i may ask to find how high above the launch point it goes.
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So my v initial, again, i'm working with vertical.
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My v initial is going to be the 26.
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My acceleration is going to be the negative 9 .8.
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This time i'm looking for the displacement.
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When i get to the highest point, at the highest point stops.
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So i'm going to say, say v final is zero.
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So i'm going to use my vs squared equation.
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Solving that for the displacement, i get 34 .4.
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And again, i'm going with two figs, so that'll be 34 meters.
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Those positives, that's how high above the launch point it is.
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And then for the third part, i'm asked to find what is a velocity when it hits the ground? well, when it hits the ground, i'm going to make my new list.
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And i'm working with vertical again.
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So my v initial is the 26.
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My acceleration is the negative 9 .8...