Asked by Alysha
A soccer ball is kicked from the ground with an initial speed of 20.4 m/s at an upward angle of 41.0˚. A player 42.2 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.
Answers
Answered by
Henry
Vo = 20.4m/s @ 41 deg.
Xo = 20.4cos41 = 15.4m/s
Yo = 20.4sin41 = 13.4m/s.
Vf^2 = Vo^2 + 2gd,
d = (Vf^2 - Vo^2) / 2g,
d(up) = (0 - (13.4)^2) / -19.6 = 6.84m,
d = Vo*t + 0.5gt^2 = 6.84m.
0 + 0.5*9.8t^2 = 6.84,
4.9t^2 = 6.84,
t^2 = 1.396,
t(down) = 1.18s.
V = d/t = 42.2 / 1.18 = 35.8m/s.
Xo = 20.4cos41 = 15.4m/s
Yo = 20.4sin41 = 13.4m/s.
Vf^2 = Vo^2 + 2gd,
d = (Vf^2 - Vo^2) / 2g,
d(up) = (0 - (13.4)^2) / -19.6 = 6.84m,
d = Vo*t + 0.5gt^2 = 6.84m.
0 + 0.5*9.8t^2 = 6.84,
4.9t^2 = 6.84,
t^2 = 1.396,
t(down) = 1.18s.
V = d/t = 42.2 / 1.18 = 35.8m/s.
Answered by
Henry
Correction:
t(up) = (Vf - Vo) / g,
t(up) = (0 - 13.4) / -9.8 = 1.37s.
T = t(up) + t(down),
T = 1.37 + 1.18 = 2.55s.
V = d/T = 42.2 / 2.55 = 16.5m/s.
t(up) = (Vf - Vo) / g,
t(up) = (0 - 13.4) / -9.8 = 1.37s.
T = t(up) + t(down),
T = 1.37 + 1.18 = 2.55s.
V = d/T = 42.2 / 2.55 = 16.5m/s.
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