Asked by stephanie
A ball is kicked at a 37 degree angle, with a velocity of 14.6 m/s. How high and far does the ball go?
Answers
Answered by
Henry
X = Hor = 14.6 m/s * cos37 = 3.67 m/s,
Y = Ver = 14.6 m/s * sin37 = 8.79 m/s,
d(up) = d(down),
8.79 m/s * t = 0.5 * 9.8 m/s^2 * t^2,
8.79t = 4.9t^2,
4.9t^2 - 8.79t = 0,
Factor out t:
t(4.9t - 8.79) = 0,
t = 0; 4.9t - 8.79 = 0, 4.9t = 8.79,
t = 8.79 / 4.9 = 1.79 s.
h = 8.79 m/s * 1.79 s = 15.76 m.
d = R = 15.76 m / sin37 = 26.2 m.
X = Hor = 26.2 cos37 = 20.9 m.
Y = Ver = 14.6 m/s * sin37 = 8.79 m/s,
d(up) = d(down),
8.79 m/s * t = 0.5 * 9.8 m/s^2 * t^2,
8.79t = 4.9t^2,
4.9t^2 - 8.79t = 0,
Factor out t:
t(4.9t - 8.79) = 0,
t = 0; 4.9t - 8.79 = 0, 4.9t = 8.79,
t = 8.79 / 4.9 = 1.79 s.
h = 8.79 m/s * 1.79 s = 15.76 m.
d = R = 15.76 m / sin37 = 26.2 m.
X = Hor = 26.2 cos37 = 20.9 m.
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