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In the javelin throw at a track-and-field event, the javelin is launched at a speed of...

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 33.0 m/s at an angle of 38.4 ° 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 38.4 ° at launch to 17.8 °?

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Answer #1

Here,

let the time taken is t

for the vertical velocity

vy = uy - g * t

vy = 33 * sin(38.4) - 9.8 * t

vy = 20.5 - 9.8 * t

for the horizontal velocity

vx = 33 * cos(38.4)

vx = 25.9 m/s

Now, for the angle to be 17.8 degree

tan(17.8 degree) = vy/vx

tan(17.8 degree) = ( 20.5 - 9.8 * t)/25.9

solving for t

t = 1.24 s

the time taken is 1.24 s

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