Javelin throw, find time it takes for angle one to change to angle two

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SUMMARY

The discussion focuses on calculating the time required for the angle of a javelin throw to change from 34° to 15° after being launched at a speed of 27 m/s. The relevant equations include the average velocity formula and the acceleration formula, with gravitational acceleration set at -9.8 m/s². The user initially attempted to derive the time using the equation t = (Vf - Vi) / a, but later found the correct answer through further calculations. The final solution demonstrates the importance of understanding both the physics concepts and the mathematical manipulation of equations.

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1. Homework Statement
In the javelin throw at a track-and-field event, the javelin is launched at a speed of 27 m/s at an angle of 34° 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 34° at launch to 15°?

Vi=27m/s
ay (g) = -9.8m/s^2
ax = 0m/s^2

2. Homework Equations
Vaverage=(Vf+Vi)/2=d/t
a=(Vf-Vi)/t

My teacher wants us to derive all other equations from this...I understand the why part of physics, it's just the "how" that I have trouble with. My teacher calls it "mathematical massaging". Also, he combines all of his equations into one giant one so that he only has to input his numbers once, no need to press "enter" several times. If you could please demonstrate how to "massage" the given equations and combine them into the larger final equation, that would be a great help.

3. The Attempt at a Solution
t=(Vf-Vi)/a
t=((cos34 deg)*(27m/s)*(sec15deg)-27m/s)/-9.8m/s^2
The reason I multiply cos34*27m/s*sec15 is so that I can find the final Velocity (cos34*27m/s gives the x component which is constant since there is no acceleration/deceleration)
my answer is .390s which isn't right (ignore sig figs, my teacher turned them off on WA for this assignment)
 
Last edited:
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never mind...I found out the answer...
 

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