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# Vertical Motion In Exercises $60-62$ , assume the accleration of the object is $a(t)=-9.8$ meters per second per second. (Neglect air resistance.)A baseball is thrown upward from a height of 2 meters with an initial velocity of 10 meters per second. Determine its maximum height.

## $$7.1 \mathrm{m}$$

Integrals

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So you know that you have to. Is it going to some starting position? Cannot, because original velocities are easier to miners. Hey, over to two squared. So this is a baseball thrown tonight, too. Cross 10 comes to my US 9.8 over to two squared and we're looking for the maximum I Now we know that the maximum i this downward facing problems I was going to occur being over to you it's gonna be over too. But this is a This is big. Hey, So, Spence way to turn We're line over to No, that's twin over. Missus Two at the max height in position. Good team Max is equal to two. Those tend turns 20 over Linus no. Or two terms. Twin over here. No squared. See Malcolm X sense

University of Delaware

Integrals

Integration

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