A batted baseball leaves the bat at an angle of 30 degrees above the horizontal and is caught by an out fielder 400ft from the plate. (a) what was the initial velocity of the ball? (b) How high did it rise? (c) How long was it in the air

Answer 1

a. Range = Vo^2*sin(2A)/g = 400 Ft.
Vo^2*sin(60)/32 = 400.
0.02706Vo^2 = 400.
Vo^2 = 14,782.
Vo = 122 Ft./s.

b. Yo = Vo*sin A = 122*sin30 = 61 Ft/s. = Ver. component of initial velocity.
Y^2 = Yo^2 + 2g*h = 0.
h = -(Yo^2)/2g = -(61^2)/-64 = 58.1 Ft.

c. Xo = Vo*Cos A=122*Cos30 = 105.7 Ft/s.
= Hor. component of initial velocity.
Range = Xo*T = 400 Ft.
105.7 * T = 400.
T = 3.78 s.