Question
A football is thrown at 14.0 m/s at an angle of 35.0 degrees. How far away does it land?
Answers
Henry
X = hor. = 14 * cos35 = 11.47 m/s.
Y = ver. = 14 * sin35 = 8.03 m/s.
d(up) = d(down),
8.03t = 0.5 * 9.8t^2,
8.03t = 4.9t^2,
4.9t^2 - 8.03t = 0,
Factor out t:
t(4.9t - 8.03) = 0,
t = 0,
4.9t - 8.03 = 0,
4.9t = 8.03,
t = 8.03 / 4.9 = 1.64 s.
d(up) = h = 8.03 m/s * 1.64 s = 13.16 m.
tan35 = h / x,
tan35 = 13.16 / x,
x = 13.16 / tan35 = 18.8 m = distance the ball lands.
Y = ver. = 14 * sin35 = 8.03 m/s.
d(up) = d(down),
8.03t = 0.5 * 9.8t^2,
8.03t = 4.9t^2,
4.9t^2 - 8.03t = 0,
Factor out t:
t(4.9t - 8.03) = 0,
t = 0,
4.9t - 8.03 = 0,
4.9t = 8.03,
t = 8.03 / 4.9 = 1.64 s.
d(up) = h = 8.03 m/s * 1.64 s = 13.16 m.
tan35 = h / x,
tan35 = 13.16 / x,
x = 13.16 / tan35 = 18.8 m = distance the ball lands.