Posted by **Erin** on Thursday, January 26, 2012 at 5:54pm.

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 35.0 m/s at an angle of 33.7 ° above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 33.7 ° at launch to 16.3 °?

- physics -
**Damon**, Thursday, January 26, 2012 at 6:01pm
Vi = 35 sin 33.7 = 19.42 initial up

u = 35 cos 33.7 = 29.12 constant horizontal

v = Vi - g t

v = 19.42 - 9.8 t

when v/u = tan 16.3 we are there

v/29.12 = .2924

v = 8.515

so

8.515 = 19.42 - 9.8 t

t = 1.11 seconds

- physics -
**drwls**, Thursday, January 26, 2012 at 6:06pm
Compute the time that it takes for the vertical velocity component Vy to decrease from 35sin33.7 to 35cos33.7*tan16.3 .

(The horizontal velocity component remains 35cos33.7)

That time will be the difference of those numbers, divided by g.

(19.420 - 8.173)/9.81 = ___ s

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