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In the javelin throw at a track and field event, the javelin is launched at a speed of 29 m/s at an angle of 35° above the hori...Asked by Sarah
In the javelin throw at a track-and-field event, the javelin is launched at a speed of 29 m/s at an angle of 36 degrees above the horizontal. As the javelin travels upward, it’s velocity point above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from 36 degrees at launch to 18 degrees?
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Answered by
oobleck
y = 29 sin36° t - 4.9t^2 = 17.045t - 4.9t^2
x = 29 cos36° t = 23.461t
at any point in flight, tanθ = dy/dx = (dy/dt) / (dx/dt)
So, you just need to find when
(17.045 - 9.8t)/23.461 = tan 18°
x = 29 cos36° t = 23.461t
at any point in flight, tanθ = dy/dx = (dy/dt) / (dx/dt)
So, you just need to find when
(17.045 - 9.8t)/23.461 = tan 18°