Two dimensional physics problem

  • Thread starter svtec
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In summary, the lower ball is thrown at a lower angle to hit the target first, and travels a shorter distance.
  • #1
svtec
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In a snowball fight 2 snowballs are thrown in the same direction, but one is thrown 70.0 degrees with respect to the horizontal and the other is thrown some arbitrary distance lower. they both are thrown with a speed of 25.0 m/s.

At what angle should the one that is thrown lower be to arrive at the same point as the one that was thrown at 70.0 degrees?

How many seconds later should the lower thrown snowball be thrown to arrive at the same time as the higher one.

resolving into x and y components for the first one i get.

x initial = 0
x final =?
v initial = 25.0cos(70.0)
v final = 0 m/s
a = 0 m/s^2

y initial = 0
y final = ?
v initial = 25.0sin(70.0)
v final = 0m/s
a = -9.80m/s^2

then i get lost...


tia...


-andrew
 
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  • #2
What makes you think "v final = 0 m/s"?
 
  • #3
i figured that when the snowball made contact with the target it's velocity would be 0.
 
  • #4
As someone who has been hit often by snowballs before, I can reassure you that as the ball hits you, it is definitely not stationary. (You are confused with collision modelling here, I think. The velocity after it hit is presumeably zero. But not as it hits/just before it hits.)

What you want to do is to use the uniform acceleration equations in each case to work out an expression for time taken in each case for the ball to fall to the ground, and then insert this to find out the horizontal distance traveled for each, and equate the two distances to find the angle.
 
  • #5
i am kinda confused. this is my first quarter in physics and i seem to confuse the moment before collision and the stationary component of the final velocity. i guess that is why i was enlisting the help of people that are far more superior in their understanding of physics then myself...

i'll give what you said a shot.

thanks...

-andrew
 

1. What is the difference between one-dimensional and two-dimensional physics problems?

One-dimensional physics problems involve motion or forces in a single direction, while two-dimensional problems involve motion or forces in two perpendicular directions. In other words, two-dimensional problems require consideration of both horizontal and vertical components, while one-dimensional problems only involve one of these components.

2. How do I solve a two-dimensional physics problem?

The first step in solving a two-dimensional physics problem is to draw a clear and accurate diagram, labeling all known and unknown quantities. Then, use vector addition to break down any forces or velocities into their horizontal and vertical components. Finally, apply the relevant equations and solve for the unknown variables.

3. Can I use the same equations for one-dimensional and two-dimensional problems?

Although some equations may be similar, two-dimensional problems require the use of vector components and may involve more complex equations. It is important to carefully consider the dimensions and components involved in the problem to determine which equations are appropriate to use.

4. What are some common mistakes when solving two-dimensional physics problems?

Some common mistakes include forgetting to consider vector components, not properly labeling the direction of forces or velocities, and not using the correct equations for two-dimensional problems. It is also important to double-check units and make sure they are consistent throughout the problem.

5. How can I check if my solution to a two-dimensional physics problem is correct?

The best way to check if your solution is correct is to make sure it satisfies the fundamental principles of physics, such as conservation of energy and momentum. You can also perform a quick estimation or use a different method to solve the problem and compare your results. Additionally, plugging in your solution to the original problem and making sure it satisfies all given conditions is another way to check for accuracy.

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