 Jesus S.

# Throwing a Ball Upwards at 12.1 m/s

You throw a baseball directly upward at time 𝑡=0 at an initial speed of 12.1 m/s. What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take 𝑔=9.80 m/s2.

By:

Jesus S.

How would we find at what times the ball passes through half the maximum height? Time is never given, is it?
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3d Mehdi R.

tutor
Y = -(g/2) (t^2) + (12.1)t → (H/2) = -4.9(t^2) +(12.1)t → -4.9(t^2) +(12.1)t - (H/2) = 0 , This is a quadratic equation which has two answers for time that passes from half maximum. One time when baseball goes up and second time, when baseball is going down from its half maximum. (t = 0.36 s and t = 2.1 s)
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2d

Jesus S.

Wow, thank you!
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2d

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