In the javelin throw at a track and field event, the javelinis launched at a spe

Question

In the javelin throw at a track and field event, the javelinis launched at a speed of 29m/s at an angle of 36o abovethe horizontal. As the javelin travels upward, its velocitypoints above the horizontal at an angle that decreases as timepasses. how much time is required for the angle to be reducedfrom 36o at launch to 18o? answer: 0.96s In the javelin throw at a track and field event, the javelinis launched at a speed of 29m/s at an angle of 36o abovethe horizontal. As the javelin travels upward, its velocitypoints above the horizontal at an angle that decreases as timepasses. how much time is required for the angle to be reducedfrom 36o at launch to 18o? answer: 0.96s

Explanation / Answer

horizontal velocity remains constant since horizontalacceleration=0 initial horizontal velocity=29*cos36 let angle is reduced to 18 degree from 36 degree after t s. let velocity after t s=v horizontal velocity after t s=v*cos18 v*cos18=29*cos36 since horizontal velocity remains constant v=24.67 m/s initial vertical velocity=29*sin36 vertical velocity after ts=24.67*sin18=7.92 v=u-g*t 24.67*sin18=29*sin36-9.8*t t=0.96 s (ans)

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