Posted by **Jack** on Sunday, January 23, 2011 at 3:14pm.

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?

- Physics -
**Henry**, Tuesday, January 25, 2011 at 5:05pm
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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