- #1

danielab87

- 10

- 0

Hi i really do not know how to do this. any help is much appreciated. thanks!

A projectile is shot out at 45° with respect to the horizon with an initial velocity of Vx = Vy = Vxy. When is the earliest time, t, that the velocity vector makes an angle of 30° with respect to the horizontal?

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ok well first i noticed that Vx=Vy=Vxy. then i decided that initially, the horizontal component of velocity is Vxycos45 and the vertical component is Vxysin45. then i thought that where the angle equals 30, the horizontal component will still be Vxycos45 but that the vertical component would be Vxysin30. then i came up with the equations: Vx=Vxycos45 and Vy=Vxysin30t-1/2gt^2. I then tried to set these equations equal to each other to solve for t but i could not get the right answer. i know that the answer is t=Vxy(1-tan30)/g, but i cannot seem to arrive at this answer. please help and tell me if I'm approaching this the right way. thanks a lot

## Homework Statement

A projectile is shot out at 45° with respect to the horizon with an initial velocity of Vx = Vy = Vxy. When is the earliest time, t, that the velocity vector makes an angle of 30° with respect to the horizontal?

## Homework Equations

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## The Attempt at a Solution

ok well first i noticed that Vx=Vy=Vxy. then i decided that initially, the horizontal component of velocity is Vxycos45 and the vertical component is Vxysin45. then i thought that where the angle equals 30, the horizontal component will still be Vxycos45 but that the vertical component would be Vxysin30. then i came up with the equations: Vx=Vxycos45 and Vy=Vxysin30t-1/2gt^2. I then tried to set these equations equal to each other to solve for t but i could not get the right answer. i know that the answer is t=Vxy(1-tan30)/g, but i cannot seem to arrive at this answer. please help and tell me if I'm approaching this the right way. thanks a lot