Asked by matt
If someone kicks a soccer ball across a flat parking lot with an initial velocity of 19.0m/s @ 19 degrees, how long will the soccer ball be in flight and how high above its launch point will the ball go? What would be the x and y components of this velocity at t=0.
Answers
Answered by
Henry
Vo = 19m/s @ 19 Deg.
Xo = 19*cos19 = 17.96 m/s.
Yo = 19*sin19 = 6.19 m/s.
Tr = (Yf-Yo)/g,
Tr = (0-6.19) / -9.8 = 0.63 s. = Rise time.
a. Tf = Tr = 0.63 s.
Tr + Tf = 0.63 + 0.63 = 1.26 s. = Time
in flight.
b. h = (Yf^2-Yo^2)/2g,
h = (0-(6.19)^2) / -19.6 = 1.95 m.
Xo = 19*cos19 = 17.96 m/s.
Yo = 19*sin19 = 6.19 m/s.
Tr = (Yf-Yo)/g,
Tr = (0-6.19) / -9.8 = 0.63 s. = Rise time.
a. Tf = Tr = 0.63 s.
Tr + Tf = 0.63 + 0.63 = 1.26 s. = Time
in flight.
b. h = (Yf^2-Yo^2)/2g,
h = (0-(6.19)^2) / -19.6 = 1.95 m.
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