- #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?
?
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
?
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