Physics Problem: Finding Final Velocity and Distance Traveled on a Ramp

  • Thread starter aerichk
  • Start date
  • Tags
    Physics
In summary, this is a two part problem involving a ball at the top of a 6.34 meter roof with a 30 degree angle at the horizontal. The goal is to find the final velocity of the ball at the end of the roof, disregarding any friction. After finding the final velocity, the task is to calculate how far the ball will travel in the air before reaching the ground. My friends and I have been struggling with this problem and any help is appreciated. The ball has a mass of 0.35kg and the gravity is 9.8m/s^2. The resultant force on the ball can be found using a cross-hair diagram and is equal to 1.715N. Using the equation
  • #1
aerichk
Ok, this is a two part problem.

Part I: We have a ball at the top of a roof that has a 30 degree angle at the horizontal. We let the ball roll and it takes 2.55 seconds to reach the bottom of the 6.34 meter roof. The ball weighs .35 kg. **DISREGARD ANY FRICTION** I.E friction of the ramp, wind resistance, any type of friction. The main part of this is to find the Final Velocity at the end of the 6.34 meter roof. So you have Vi(initial velocity) = 0, Vf(Final Velocity) = ?, G(gravity) = 9.8m/s^2. The ramp has a 30 degree angle at the horizontal, and it takes 2.55 seconds to get to the bottom.

Part II: After we find the Final Velocity, we have to figure out how far the ball will travel in the air. After it reaches the final velocity, it falls off the roof 2.51 meters to the ground. The *main part* of the entire problem is to figure out how far it will go off the roof. The Y axis is 2.51 meters, and we need to find out what the X axis will be. We need to find out Dx(distance of x), Vx(Velocity of x), and T(time it takes to reach the ground).

My friends and I have been having trouble with this problem all day. We keep coming up with wrong, impossible answers. Any help is appreciated. Thanks.
 
Physics news on Phys.org
  • #2
First of all a ball cannot weigh .35kg, it has a mass of .35kg. Thats like saying the distance from A to B is 57Volts.

The first step to solving ur problem is finding the resultant force acting on the ball. This can be done by drawing one of those cross-hair things. I will do my best to demonstrate how, using only characters on the ASCII pallette.

*
*
o*
...
| ...
| ...
| ...

... = roof
o = ball
| = Force of weight =mass x accel due to grav= 0.35 x 9.8 = 3.43
* = The normal reaction force of the roof (acting perpendicular to the roof)

If you place a crosshair on the system so the the middle is on the body and one line is parallel to the normal force of the roof and the other is offset 30 degrees clockwise to the force of weight.

You can say that the forces acting perpendicular to the roof are ballanced since there is no change in velocity in a perpendicular direction with respect to the roof. Taking the perpendicular componant of the Weight using trig would be 3.43Cos30 and therefore the normal force would be 3.43Cos30. This means that the resulatant force ends up going down the roof 3.43Sin30 = 1.175N

Put that into F=ma (resultant force = mass x accel) to get

1.715 = 0.35a therefore a = 0.6ms^-2 down the roof

substitute this value into v=u+at to get v=0.6 x2.55 = 1.53

DO THE REST UR SELF. Maybe i have helped u enuf to do so

it has been an honour to answer ur question. If u need more help chat 2 me on msn on mceddy2001@hotmail.com. Soz about the blabber but i can't draw the diagrams on here so i need 2 explain more. If u want me 2 draw a better diagram in paint i will be happy providing i have the time
 
  • #3
sorry my diagram which i spend ages on got ruined by the system. Many apologies it seems to have taken out all of the necessary spaces.
 

1. What is the definition of "Easy Physics problem"?

"Easy Physics problem" refers to a physics problem that is relatively simple to solve and does not require advanced mathematical or conceptual understanding. These types of problems often involve basic equations and concepts, making them accessible to those new to physics.

2. How can I identify an "Easy Physics problem"?

An "Easy Physics problem" can typically be identified by the use of familiar equations, such as Newton's laws or the equations of motion. These problems often involve simple calculations and do not require an in-depth understanding of advanced concepts.

3. Are "Easy Physics problems" important to study?

Yes, "Easy Physics problems" are important to study as they provide a foundation for understanding more complex concepts and theories. By mastering these basic problems, students can build their problem-solving skills and gain a better understanding of fundamental physics principles.

4. How can I improve my ability to solve "Easy Physics problems"?

To improve your ability to solve "Easy Physics problems", it is important to practice regularly and familiarize yourself with common equations and concepts. Additionally, seeking help from a teacher or tutor and working through example problems can also improve your problem-solving skills.

5. Can "Easy Physics problems" have real-world applications?

Yes, "Easy Physics problems" often have real-world applications, particularly in fields such as engineering, where basic physics principles are used to solve practical problems. Understanding these problems can also help in everyday situations, such as calculating the speed or distance of an object.

Similar threads

  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
783
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
952
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
2
Replies
40
Views
2K
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
847
Back
Top