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University of Houston

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Problem 61

With what initial velocity must an object be thrown upward (from a height of 2 meters) to reach a maximum height of 200 meters?

Answer

$$

v_{0} \approx 62.3 \mathrm{m} / \mathrm{s}^{2}

$$

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## Discussion

## Video Transcript

era in question 51 we will be using the function from from question 60 which is F A T equals negative board White nine. He squared close our mission velocity time. Tea glass too. I was recounting question 60 though I am going to take the derivative of that because we want the reach Max of 200. So I'm taking derivatives. Um, so negative. 4.9 times to gives me negative 9.8 Jeanne plus these of zero and now been substitutes he ruin for my pride and solved. And now, in my position, equation of good. Couldn't my cheese place my p So I have Yeah. Oh, over point No. 19 from eight. I swear. These three ground times Gypsy rose, divided by plus two equals two *** or quite nine. The Pepsi roads where over 98 subzero square eight equal venom is attracted to from both sides on 98. And now I have multiple other. Multiply everything by 9.8 square. You get rid of the denominator, have negative or 0.9 visa zeros. Where lies the square with zero eyes? 9.8 Because the dominator only has one night, but need, innit? People's 1 98 I'm 9.8 square. So now I have, like, terms that I can combine that native 4.9 beasts Shapiro's where plus 9.8, These zeros Where gives me positive or 0.9. These have zeros Square. And this e equal owns one. Maybe one 19 0 15 White 92. We're gonna divide that by 4.9 and that's gonna give me 38 8 equals V sub zero's where Take this. We're route of that and I get 623 meters per second equals my initial no.

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