How long does it take for the angle of the javelin to change from 35 to 20?

[Ohoneo]

Homework Statement

In the javelin throw at a track-and-field event, the javelin is launched at a speed of 26 m/s at an angle of 35° 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 35° at launch to 20°?

Homework Equations

vy = v0y + ay t
y = x0 + vy t
y = v0y t + 1/2 ay t^2
vy^2 = v0y^2 + 2 ay y
(And also substitute x for y in the above for horizontal components)
vx = v cos theta
vy = v sin theta

The Attempt at a Solution

Well, I thought that if I could find out how much the y component changes, I could plug it into the vy = v0y + ay t equation.
So, I found the vertical component of when the javelin is released at 35 degrees to be 15 m/s, and the vertical component at 20 degrees is 8.9 m/s.
I found this by:
vy = 26 sin 35 = 15 m/s
vy = 26 sin 20 = 8.9 m/s
Then, I plugged it into the kinematic equation:
8.9 m = 15 m + (-9.8 m/s)t
t = .62 seconds

Unfortunately, this turned out to be incorrect. Any help is appreciated! :)

 

[tiny-tim]

Hi Ohoneo!

(have a theta: θ and try using the X² and X₂ icons just above the Reply box )

vy = 26 sin 35 = 15 m/s
vy = 26 sin 20 = 8.9 m/s

No, you're assuming that the speed is constant.

What is constant?

 

[Ohoneo]

Hi Ohoneo!

(have a theta: θ and try using the X² and X₂ icons just above the Reply box )


No, you're assuming that the speed is constant.

What is constant?

Ah, thanks! I was wondering how to enter those!
See, that's where I'm confused. Acceleration would be constant in the y direction (-9.80 m/s²), and then acceleration would be 0 in the x direction... but, then, nothing else is constant.

 

[tiny-tim]

… acceleration would be 0 in the x direction...

yup! … so the component of velocity in the x direction must be constant!

 

[Ohoneo]

yup! … so the component of velocity in the x direction must be constant!

Alright, so using that I found that at 20 degrees, v_x = 21 m/s
So, I need to find v, or v_y.
to do that, I plugged in 21 = v cos 20
So I found that v = 22 m/s
Then I plugged that into 22² = 21² + v_y²
So I got v_y = 7.6. So I plugged into v_y(final) = v_y(initial) + a_yt
And I got t = .13 seconds.. which is still wrong =/

 

[tiny-tim]

Hi Ohoneo!

Your equations look correct, but you're not using enough significant figures …

in the middle of a calculation, you must always use more sig figs than you need in the final answer …

when you subtract 21² from 22², you obviously need the 22 to be a lot more accurate than to the nearest whole number!

Try again, using two decimal places until the end. ​