How Is Time Calculated for Angle Reduction in Javelin Throw Projectile Motion?

[sweetlavender]

4.In the javelin throw at a track-and-field event, the javelin is launched at a speed of 30 m/s at an angle of 33° 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 33° at launch to 15°?
i don't understand how we can find the difference between the time without them giving the velocity at 15degrees?

 

[Kurdt]

You will have to split the motion into 2 components. The horizontal and the vertical. The vertical is subject to the acceleration due to gravity and the horizontal should be assumed to be constant. You need to find the vertical component that gives a resultant that has an angle of 15 degrees with the horizontal. Then you can find the time it takes to go from the original vertical component of velocity to the one at 15 degrees.

 

[sweetlavender]

yes i did get the components of x and y
Vox=(30m/s)cos33=25.16m/s
Voy=(30m/s)sin33=16.339m/s
and then i tried to find the total time it takes with t=-2(Voy/g). that's it i am stuck over there i don't now how to go further :(

 

[lightgrav]

did you find the vertical component to get 15 degrees?
You know, of course, that the horiz.velocity doesn't change
...so you know 1 side and 1 angle (as well as the 90)