Posted by Anonymous on .
A ball is thrown with an initial velocity of 20 meters per second at an angle of 60° with the horizontal ground surface. Find the following:
a. the horizontal velocity of the ball
b. the ball’s initial vertical velocity
c. the range of the ball
d. the maximum height of the ball
Vo = 20 m/s @ 60 Deg.
a. Xo = 20*cos60 = 10 m/s.
b. Yo = 20*sin60 = 17.3 m/s.
c. Tr = (Vf - Vo) / g,
Tr = (0 - 17.3) / -9.8 = 1.77 s. = Rise
Tf = Tr = 1.77 s.
Tr + Tf = Time in flight.
R = Xo * (Tr+Tf),
R = 10 m/s. * (1.77+1.77) s. = 35.4 m.
d. h max = (Yf^2 - Yo^2) / 2g,
h max = (0 - (17.3)^2) / -19.6=15.31 m