Projectile Problem: Marble falls from Desk

In summary, the conversation revolved around a question related to a marble that rolls off a desk and falls to the floor. The conversation covered various aspects of the question, such as the marble's horizontal and vertical velocities at different points during the fall, and the correct way to represent the acceleration due to gravity in a vector equation. The final conclusion was that the marble's vertical velocity increases as it falls, and that the diagram provided by the OP was clear and sufficient for understanding the problem.
  • #1
aatari
73
3
Hey Guys, I have attempted the question below and need someone to check my work, please.

Thanks!

1. Homework Statement

A marble rolls off a desk that is 0.86 m above the floor with a horizontal speed of 1.2 m/s.

a) What is the marble's horizontal velocity when it is half way to the floor?
b) What is the marble's vertical velocity at that point?
c) Describe the value of the vertical component of velocity as the marble falls to the floor.​

Homework Equations

The Attempt at a Solution


20170224_173634[545].jpg
 
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  • #2
For c), I don't think that ##0## is the answer they are looking for!
 
  • #3
Also, if you are using ##g## to one decimal place, you cannot have an answer for velocity to four decimal places.
 
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  • #4
PeroK said:
For c), I don't think that ##0## is the answer they are looking for!
Will it be 2.9008 m/s?
 
  • #5
It is not correct to write ##\vec g = - 9.8 m/s^2## , because a vector is never equal to a number.

After the ball leaves the table, the vertical component of its speed increases from ##9.8 m/s## to every second until it collides with the ground.
 
Last edited:
  • #6
aatari said:
Will it be 2.9008 m/s?

No, you just do what you did to find its velocity half way to the floor.

You can safely ignore post #5.
 
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  • #7
PeroK said:
No, you just do what you did to find its velocity half way to the floor.

You can safely ignore post #5.
I get it, thank you.
It will be 4.1 m/s.
 
  • #8
aatari said:
I get it, thank you.
It will be 4.1 m/s.

That looks about right.
 
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  • #9
aatari said:
I get it, thank you.
It will be 4.1 m/s.
In letter c he does not ask for speed by hitting the ground. He asks what happens to the speed over the fall.
 
  • #10
Caio Graco said:
In letter c he does not ask for speed by hitting the ground. He asks what happens to the speed over the fall.
Good point.
 
  • #11
Caio Graco said:
In letter c he does not ask for speed by hitting the ground. He asks what happens to the speed over the fall.
For this, see post # 5 "ignored".
 
  • #12
aatari said:
I get it, thank you.
It will be 4.1 m/s.

It looks like they weren't asking for the speed as it hits the floor, but a "description of the value" during the fall.
 
  • #13
PeroK said:
It looks like they weren't asking for the speed as it hits the floor, but a "description of the value" during the fall.
Exactly
 
  • #14
Caio Graco said:
For this, see post # 5 "ignored".

The thing to ignore was about the vector. The OP's diagram is good and clear in its vector representation, in my opinion.
 
  • #15
PeroK said:
The thing to ignore was about the vector. The OP's diagram is good and clear in its vector representation, in my opinion.
The problem is that he wrote ##\vec g= - 9.8 m/s^2##
The correct is ##\vec g= -9.8 \vec j m/s^2## (vector equation) or else I could write ##g=-9.8 m/s^2## (scalar equation)
 
  • #16
Caio Graco said:
The problem is that he wrote ##\vec g= - 9.8 m/s^2##
The correct is ##\vec g= -9.8 \vec j m/s^2## (vector equation) or else I could write ##g=-9.8 m/s^2## (scalar equation)

The OP's diagram is clear and that's just pointless pedantry.
 
  • #17
Caio Graco said:
g=−9.8m/s2
When writing this it is being said that a vector is equal to a number. A vector can only be a vector and a number can only equal a number.
 
  • #18
PeroK said:
The OP's diagram is clear and that's just pointless pedantry.
Ok.
 
  • #19
Caio Graco said:
When writing this it is being said that a vector is equal to a number. A vector can only be a vector and a number can only equal a number.
Yes, you've made your point, but it really is a bit excessive to complain about that in an explanatory note in a diagram.
 
  • #20
haruspex said:
Yes, you've made your point, but it really is a bit excessive to complain about that in an explanatory note in a diagram.
I agree. It's just a matter of formality.
 

1. How do you calculate the time it takes for the marble to hit the ground?

The time it takes for the marble to hit the ground can be calculated using the formula t = √(2h/g), where t is the time in seconds, h is the height of the desk in meters, and g is the acceleration due to gravity (9.8 m/s²).

2. What is the initial velocity of the marble when it falls from the desk?

The initial velocity of the marble is equal to 0 m/s since it is not given any initial force or velocity when it falls from the desk.

3. How far does the marble travel horizontally before hitting the ground?

The horizontal distance traveled by the marble before hitting the ground can be calculated using the formula d = v*t, where d is the distance in meters, v is the horizontal velocity of the marble (0 m/s), and t is the time in seconds.

4. Is the air resistance taken into account when solving this projectile problem?

In most cases, air resistance is neglected when solving projectile problems since it is typically small and does not have a significant effect on the results. However, in more complex scenarios, air resistance can be taken into account by modifying the equations used for calculation.

5. Can the height of the desk be changed to affect the time it takes for the marble to hit the ground?

Yes, the height of the desk directly affects the time it takes for the marble to hit the ground. As the height increases, the time also increases since the marble has more distance to cover before reaching the ground. This relationship is described by the formula t = √(2h/g), where t is the time in seconds, h is the height of the desk in meters, and g is the acceleration due to gravity (9.8 m/s²).

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