Physics
posted by Jack .
A golf ball is chipped at an angle of 49 degrees and with a speed of 8.5 m/s. How far does it travel to the nearest tenth of a meter?

X = hor = 8.5*cos49 = 5.58m/s.
Y = ver = 8.5*sin49 = 6.42m/s.
The velocity = 0 at the max height:
Vf = Vi + gt = o,
6.42 + (9.8)t = 0,
6.42  9.8t = o,
9.8t = 6.42,
t(up) = 6.42 / 9.8 = 0.66s.
t(tot.) = 2 * 0.66 = 1.32s.
d = 5.58m/s * 1.32s = 7.4m.
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