A ball is kicked off a field with sufficient force and a trajectory such that it travels horizontally 60m with a maximum altitude of 20m.
What was the initial speed?
What was the trajectory angle?
a. h = Vo*t + 0.5g*t^2 = 20.
0 + 4.9t^2 = 20
t^2 = 4.08
Tf = 2.02 s. = Fall time.
Tr = Tf = 2.02 s. = Rise time
Dx = Xo(Tr+Tf) = 60 m.
Xo * 4.04 = 60
Xo=14.85 m/s.=Hor. component of initial velocity.
Y = Yo + g*Tr = 0.
Yo - 9.8*2.02 = 0
Yo - 19.8 = 0
Yo = 19.8 m/s = Ver. component of initial velocity.
a. Vo^2 = Xo^2 + Yo^2.
V0^2 = 14.85)^2 + (19.8)^2 = 612.56
Vo = 24.75 m/s = Initial velocity.
b. tanA = Yo/Xo = 19.8/14.85 = 1.33333
A = 53.1o