Question
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
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
Answers
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
time.
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.
= Range.
d. h max = (Yf^2 - Yo^2) / 2g,
h max = (0 - (17.3)^2) / -19.6=15.31 m
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
time.
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.
= Range.
d. h max = (Yf^2 - Yo^2) / 2g,
h max = (0 - (17.3)^2) / -19.6=15.31 m
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