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Question 76 states that a men's world record for the shot put, 23 .12 meters, was set by randy barnes of the united states on may 20th, 1990.
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If the shot was launched from six feet above the ground at an initial angle of 42 degrees, what is its initial speed? all right, so we have a shot putter that launches it at a distance i called the y of six feet off the ground, but i converted to 1 .83 meters, because that's what we have.
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We'd like to see.
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It's launched, you know, the horizontal distance, dx is 23 meters.
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We're looking to see what that initial launch speed was of v -0.
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If it's launched, assumingly at some angle, well, not assumingly, sorry, tells us 42 degrees.
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Yeah, what is v -0? okay, so we're told, again, all these terms, dx, d, d, y, and theta.
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So what else may not be obvious to start.
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I'm just going to pick one of the equations that has all of these variables.
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So i'm going to start with my dx equation.
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So my dx, i know that's related to my initial velocity by, you know, v0 times t.
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Well, okay, well, how can this help me? well, i can say, okay, so that's my initial velocity.
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If i took my angle here, if this is theta and this is v0, then my horizontal component or the x direction, which, hey, i should label my x direction, what that means.
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That would be correspond to the cosine of my v0, so that's v0, cosine theta times time and now i'm stuck right i don't know v not i don't know time but i need to solve for v not so i need to keep that term there so perhaps if i first i'll look at my dy equation if i look at this i see okay if i can replace my t term with with what i can get for my x term equation i can eliminate time as a variable and only have v not as my unknown variable so i'm going to do that and note for with my d -y equation, i recognize that my distance in the y direction is negative because i'm going from a higher height to a lower height.
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And of course, the acceleration due to gravity is zero, or is negative as well, not zero.
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And our initial launch speed will have a positive y value.
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So that's why my negatives are real on those two terms.
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And of course, is a t -square at the end here.
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So like i said, in this equation, i'm going to solve for time so i can eliminate that unknown variable and only have b -0 as my unknown.
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So that's my plan here.
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So v .0 cosine theta.
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I can now substitute that into my d .y equation.
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So if i have negative dy, my initial velocity in the y direction is simply v0...