A weird projectile problem

In summary: Vx(t)=Vy(t)*tan 45, but since Vx=Vy, Vx(t)=Vx=Vy(t)*tan 45=Vy(t)combining with the above eqn, t=Vx/(g*tan 45) (2)since (1)=(2), we can cross multiply and solve for t, t=Vx/(g*tan 45)=(Vy/(2*g))*(1/tan 30) or t=Vx/(g*tan 45)=(Vy/(2*g))*(Vx/Vy) or t=Vx/(g*tan 45)=1/2.
  • #1
danielab87
10
0
Hi i really do not know how to do this. any help is much appreciated. thanks!

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
 

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  • #2
Okay, this is just a standard trig problem. since there is no forces affecting the horizontal plane, velocity in the x is a constant. if you wana know when the angle is 30, and you have a constant side. so ( tan(30) ) = ( Vy / Vx ). and the other problem with your equasion. "Vy=Vxysin30t-1/2gt^2" is false. Y=Vxysin30t-1/2gt^2 take note that its equal to 'Y' (the distance) your looking for the ratio of the velocities.
 
  • #3
oh my fault i meant to put Vy=Vxysin30-gt

i don't know why i put that
 
  • #4
so you have a soln?
 
  • #5
i don't think this part of the data is right...it seems incorrect to me...
[tex] v(x) = v(y) = v(xy) [/tex]

if just [tex] v(x) = v(y) [/tex]

things may be a lot simpler...
 
  • #6
Well Vx=Vy. Usually the total velocity gets called something different than Vxy. Vxy if total can't be equal to either Vx or Vy. By virtue of 45 degrees, Vx does equal Vy, but nothing more.
 
  • #7
I agree that Vxy can't equal Vx or Vy...however, I do have somewhat of a solution (Kinda sketchy :p):

We know that wherever this 30 degrees point is, the Vx will be the same as the horizontal so it can be denoted Vy since Vx = Vy. To find the y component, it would be tan30*Vy (Soh Cah toa). Then you know the acceleration is g so we can use:
g= Dv/Dt ---> t= Dv/g ---> t= (V2-V1)/g ---> t= (Vytan30 - Vy)/g -->
t= Vy(tan30 -1)/g -----sub Vxy for Vy since they are equal(?) --> t=Vxy(tan30-1)/g
 
  • #8
looks good: the way I solved it was similar;

Vy(t)=Vy(init)-g*t
Vx(t)=contant=Vx(init)

at 30 degrees, Vy(t)/Vx(t)=tan 30=1/2, and dividing the above eqns,

1/2=1-g*t/Vy or t=Vy/(2*g)
 

1. What is a weird projectile problem?

A weird projectile problem is a physics problem that involves the motion of a projectile (an object that is launched into the air) in a strange or unusual way. This could involve factors such as non-uniform gravity, air resistance, or a curved trajectory.

2. How do you solve a weird projectile problem?

The best way to solve a weird projectile problem is to break it down into smaller, more manageable parts. Start by identifying the initial conditions, such as the launch angle, initial velocity, and any other relevant factors. Then, use equations of motion and principles of physics to analyze the motion of the projectile and solve for the desired variables.

3. What are some real-life examples of weird projectile problems?

Some real-life examples of weird projectile problems include the motion of a baseball thrown with spin, the trajectory of a rocket launch, and the flight path of a frisbee thrown on a windy day. These scenarios involve factors that make the projectile's motion more complex and unpredictable.

4. How do you account for air resistance in a weird projectile problem?

To account for air resistance in a weird projectile problem, you can use the drag force equation, which takes into account the object's velocity, size, and shape. This force can then be incorporated into the equations of motion to more accurately predict the projectile's trajectory.

5. Are there any real-world applications for solving weird projectile problems?

Yes, there are many real-world applications for solving weird projectile problems. Engineers and scientists use these principles to design and improve objects such as airplanes, rockets, and sports equipment. Understanding the motion of projectiles can also help in predicting the path of natural phenomena, such as volcanic eruptions or asteroid impacts.

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