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western50 · Dec 3, 2010, 01:04 AM

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))

Unknown008 · Dec 3, 2010, 01:40 AM

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.

ebaines · Dec 3, 2010, 11:25 AM

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

$<br /> v'_y = v_y - gt \\<br /> \frac {v_y} {\sqrt 3} = v_y - gt<br />$

Unknown008 · Dec 3, 2010, 11:27 AM

Oh, duh, thanks, I missed that :o

I'll edit my post and correct it now.