https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys100/fall10/preflights/13/fa04_1.gif

A projectile is shot out at 45o with respect to the horizontal with an initial velocity of vx = vy = vxy. When is the earliest time, t, that the velocity vector makes an angle of 30o with respect to the horizontal?
t = vxy / (g sin(30o))
t = vxy(1 - sin(30o)) / g
t = vxytan(30o) / g
t = vxy(1 - tan(30o)) / g
t = vxy / (g tan(30o))

I would do it as such:

When v_y = v_x, then angle is given by

\tan^{-1}(\frac{v_y}{v_x}) = 45^o

When the angle is 30, we get:

\tan(30) = \frac{1}{\sqrt3}

Hence, the ratio of the new v'y and vx is:

\frac{v'_y}{v_x} = \frac{1}{\sqrt3}

when the angle is 30 degrees.

Cross multiply, and since vx = vy;

\frac{v_y}{\sqrt3} = v'_y

When does v_y becomes v'_y?

Using v = u + at

v'_y= v_y - gt

\frac{v_y}{\sqrt3}= v_y - gt

EDIT: Mistake on the direction of g.

Unk: one minor correction: I believe that since the acceleration in the vertical direction is equal to -g, the last two lines should be:


v'_y = v_y - gt \\
\frac {v_y} {\sqrt 3} = v_y - gt

Oh, duh, thanks, I missed that :o

I'll edit my post and correct it now.