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## Homework Statement

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 26 m/s at an angle of 35° 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 35° at launch to 20°?

## Homework Equations

vy = v0y + ay t

y = x0 + vy t

y = v0y t + 1/2 ay t^2

vy^2 = v0y^2 + 2 ay y

(And also substitute x for y in the above for horizontal components)

vx = v cos theta

vy = v sin theta

## The Attempt at a Solution

Well, I thought that if I could find out how much the y component changes, I could plug it into the vy = v0y + ay t equation.

So, I found the vertical component of when the javelin is released at 35 degrees to be 15 m/s, and the vertical component at 20 degrees is 8.9 m/s.

I found this by:

vy = 26 sin 35 = 15 m/s

vy = 26 sin 20 = 8.9 m/s

Then, I plugged it into the kinematic equation:

8.9 m = 15 m + (-9.8 m/s)t

t = .62 seconds

Unfortunately, this turned out to be incorrect. Any help is appreciated! :)