posted by deb on .
major leaguer hits a baseball so that it leaves the bat at a speed of 29.0m/s and the angles of 36.9 above horizontal ignore resistance what two time is the baseball at a height of 10.0m above the point at which it left the bat?
*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.
Vo = 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.