Sunday

February 1, 2015

February 1, 2015

Posted by **deb** on Saturday, September 28, 2013 at 1:54am.

*what two times is the baseball at a height of 10.0m above the point at which it left the bat?

Calculate the horizontal component of the baseballs velocity at each of the two times you found in part a.

calculate the vertical component of the baseball's velocity at each of the two times you found in part (a)?

What are the magnitude of the baseball's velocity when it returns t the level at which it left the bat?

Thanks for any help, when I calculate everything I'm missing a step and the answer is wrong.

- physics -
**Henry**, Sunday, September 29, 2013 at 10:07pmVo = 29m/s[36.9o]

Xo = 29*cos36.9 = 23.2 m/s.

Yo = 29*sin36.9 = 17.4 m/s.

a. h = Yo*t + 0.5g*t^2 = 10 m

17.4*t - 4.9t^2 = 10

-4.9t^2 + 17.4t - 10 = 0. Use Quad. Formula.

t = 0.72, and 2.83 s.

b. X = Xo = 23.2 m/s. = Hor. component

of velocity , and does not change.

c. Y = Yo + g*t = 17.4 - 9.8*0.72 = 10.34 m/s.

Y = 17.4 - 9.8*2.83 = -10.34 m/s.

d. V = Vo = 29 m/s.

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