In the javelin throw at a track-and-field event, the javelin is launched at a sp

Question

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 25.0 m/s at an angle of 30.2 ° 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 30.2 ° at launch to 10.7 °? In the javelin throw at a track-and-field event, the javelin is launched at a speed of 25.0 m/s at an angle of 30.2 ° 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 30.2 ° at launch to 10.7 °? In the javelin throw at a track-and-field event, the javelin is launched at a speed of 25.0 m/s at an angle of 30.2 ° 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 30.2 ° at launch to 10.7 °?

Explanation / Answer

the angle that the velocity vector makes with the horizontal is given by
tan = Vy/Vx

where,

Vy = V0y + ayt
= V0sin0 + ayt

So,

t = V0 (cos0 tan - sin0) / ay

= (25 m/s)(cos30.2 *tan10.7 - sin30.2) / -9.8 m/s^2

= 0.866 s

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