posted by Gee .
A 3rd baseman is throwing a ball to 1st base which is 39.0m away. It leaves his hand at 38.0m/s at a height of 1.50m from the ground and makes an angle of 20 Degrees. How high will it be when it gets to 1st base.
Vo = 38 m/s @ 20o.
Xo = 38*cos20 = 35.71 m/s.
Yo = 38*sin20 = 13.0 m/s.
Xo * Tr = 39 m.
35.71*Tr = 39
Tr = 1.09 s. = Time to reach 39 meters.
h = ho + Yo*Tr + 0.5g*Tr^2.
h = 1.5 + (13*1.09 - 4.9(1.090)^2)=9.85
Meters above gnd when it reaches 1st base.