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How far does the ball go down the ramp

  • Thread starter Entr0py
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102
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1. Homework Statement
A ball is shot out of a gun with an initial speed of 110 m/s. The acceleration of ball is -6 m/s^2 as it goes up an inclined plane of 1 m, which makes a 30 degree angle with the horizontal. Once the ball reaches the top of the inclined plane, it is launched off onto another inclined plane, which makes a 30 degree angle with the horizontal. These are the questions: what is the speed of the ball the moment it is released from first inclined plane. How far does the ball go before it hits the second inclined plane? What is the time it takes the ball to hit the second inclined plane? What are the coordinates of the point where ball hits the second inclined plane?

2. Homework Equations
I am very confused which equations I am able to use. I could find the ball's speed the moment if flies off the first inclined plane. I know the formula for that: V^2=V0^2+2ax. But all other formulas evade me at the moment.

3. The Attempt at a Solution
This was a difficult problem that my professor couldn't even solve. I'm very determined to solve this problem as it is very awesome.
 
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andrewkirk

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First, I assume you mean the ball's acceleration while on the first ramp is -6##ms^{-2}##, not -6m/s.

Calculating the ball's velocity when it leaves the first ramp is easy and it sounds like you can do that.

The problem cannot be solved beyond that without more info, specifically:

1. what is the acceleration due to gravity?
2. what is the equation of the second plane (say in a coordinate system in which the top of the first plane has coordinates (0,0) and the y axis points in the opposite direction from gravity)?

Normally we could assume for 1 that g=9.8##ms^{-2}## but we can't do that here because that would give a deceleration of -5##ms^{-2}## on the first ramp, which is different from what was given. That could be explained either by gravity being different or by a frictional deceleration of -1##ms^{-2}## on the first ramp. We don't know which is applicable.

Once 1 and 2 are cleared up, the problem is easy. It's just finding the intersection between a straight line and a parabola, which just involves a bit of manipulation and solving a quadratic.
 
102
1
First, I assume you mean the ball's acceleration while on the first ramp is -6##ms^{-2}##, not -6m/s.

Calculating the ball's velocity when it leaves the first ramp is easy and it sounds like you can do that.

The problem cannot be solved beyond that without more info, specifically:

1. what is the acceleration due to gravity?
2. what is the equation of the second plane (say in a coordinate system in which the top of the first plane has coordinates (0,0) and the y axis points in the opposite direction from gravity)?

Normally we could assume for 1 that g=9.8##ms^{-2}## but we can't do that here because that would give a deceleration of -5##ms^{-2}## on the first ramp, which is different from what was given. That could be explained either by gravity being different or by a frictional deceleration of -1##ms^{-2}## on the first ramp. We don't know which is applicable.

Once 1 and 2 are cleared up, the problem is easy. It's just finding the intersection between a straight line and a parabola, which just involves a bit of manipulation and solving a quadratic.
Yes I meant the acceleration is -6 m/s^2. Could you clarify what you mean by "manipulating a quadratic"? Also, why would we have to know the intersection of a straight line and quadratic? Thanks
 

andrewkirk

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The flight path of the ball will be a parabola. The surface of the second ramp is a straight line. The problem asks for information about where the two intersect.
 

Mister T

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I think it's safe to assume that ##g=9.8\ \mathrm{m/s^2}## and that there is friction on the incline. When the ball reaches the end of the first inclined plane it becomes a projectile. The other assumption I would make is that the second inclined plane has the same height as the first.

If your instructor didn't give this problem to you, where did you get it?
 
102
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I think it's safe to assume that ##g=9.8\ \mathrm{m/s^2}## and that there is friction on the incline. When the ball reaches the end of the first inclined plane it becomes a projectile. The other assumption I would make is that the second inclined plane has the same height as the first.

If your instructor didn't give this problem to you, where did you get it?
He did give it to our class. However, he couldn't solve it
 

Mister T

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He did give it to our class. However, he couldn't solve it
My guess is that a figure is supposed to accompany the problem. Without that figure, it can't be solved. It's a projectile problem.

In your first post you said you were able to calculate the speed at the top of the first ramp. That's the launch speed of the projectile. The launch angle is 30°. Do you know how to proceed from here?

Ramps.png
 

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