What Angle Below the Horizontal Does a Hockey Puck Hit the Ground?

In summary, the puck slides off the edge of a table and has an initial vertical velocity of 20 m/s. It takes 0.63 seconds for it to hit the ground.
  • #1
Im_eNVy
7
0

Homework Statement


A hockey puck slides off the edge of a table with an initial velocity of 20 m/s. The height of the table above the ground is 2 m. At what angle below the horizontal does the puck hit the ground?

Homework Equations


d=Vi(t)+(1/2)at^2
a^2+b^2=c^2

The Attempt at a Solution


Horizontal Values: Vi=20Cos(theta), a=0, Xi=0m, Xf=?
Vertical Values: Vi=20sin(theta), a=-9.81m/s^2, Xi=2m, Xf=0

1) 2=20sin(theta)t+(0.5)(-9.81)t^2
2)1/10=sin(theta)t+(-4.905)t^2
3) 1/10 x 1/(-4.905)=sin(theta)t+t^2

Stuck after solving that part.
 
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  • #2
Im_eNVy said:

Homework Statement


A hockey puck slides off the edge of a table with an initial velocity of 20 m/s. The height of the table above the ground is 2 m. At what angle below the horizontal does the puck hit the ground?

Homework Equations


d=Vi(t)+(1/2)at^2
a^2+b^2=c^2

The Attempt at a Solution


Horizontal Values: Vi=20Cos(theta), a=0, Xi=0m, Xf=?
Vertical Values: Vi=20sin(theta), a=-9.81m/s^2, Xi=2m, Xf=0

1) 2=20sin(theta)t+(0.5)(-9.81)t^2
2)1/10=sin(theta)t+(-4.905)t^2
3) 1/10 x 1/(-4.905)=sin(theta)t+t^2

Stuck after solving that part.
The puck is initially 2 m above the ground when it slides off the table. How long does it take for the puck to hit the ground?

Hint: the amount of time it takes the puck to fall is not affected by its horizontal velocity.
 
  • #3
I thought there is a horizontal velocity? I'm unsure what you mean, if I use the horizontal value then I do not have the displacement for it. I'm attempting to find the time using the vertical initial velocity, but i have two missing variables: theta and t. I can't do a quadratic. I'm unsure if there even is an initial vertical velocity.
 
  • #4
Wait, do I have my initial velocities backwards?
 
  • #5
Im_eNVy said:
I thought there is a horizontal velocity? I'm unsure what you mean, if I use the horizontal value then I do not have the displacement for it. I'm attempting to find the time using the vertical initial velocity, but i have two missing variables: theta and t. I can't do a quadratic. I'm unsure if there even is an initial vertical velocity.
I never said there wasn't a horizontal velocity, because the puck comes flying off the table at 20 m/s.

What is important to remember is, the puck takes the same amount of time to fall 2 m whether it comes flying off the table at 20 m/s or 200 m/s. The puck is coming off the table horizontally, so there is no initial upward velocity component for gravity to overcome. If there is no initial upward velocity component for gravity to overcome, then what is the puck's initial vertical velocity?

If you can't "do" a quadratic, there is a handy formula called, surprisingly enough, the "quadratic formula" which can get you over that hump.
 
  • #6
SteamKing said:
I never said there wasn't a horizontal velocity, because the puck comes flying off the table at 20 m/s.

What is important to remember is, the puck takes the same amount of time to fall 2 m whether it comes flying off the table at 20 m/s or 200 m/s. The puck is coming off the table horizontally, so there is no initial upward velocity component for gravity to overcome. If there is no initial upward velocity component for gravity to overcome, then what is the puck's initial vertical velocity?

If you can't "do" a quadratic, there is a handy formula called, surprisingly enough, the "quadratic formula" which can get you over that hump.

I meant I did not have the values required to do the quadratic, no need for the ignorant comment. There's what I wanted to know, initial vertical velocity is 0m/s. Thank you, I get confused with free fall off a platform and projectiles.
 
  • #7
I'm stuck again, frustrating to be stuck on a simple question like this. Time calculated to be 0.63s

d=20cos(theta)(0.63)

I'm missing 2 values, how do i solve it from here on?
 
  • #8
Im_eNVy said:
I'm stuck again, frustrating to be stuck on a simple question like this. Time calculated to be 0.63s

d=20cos(theta)(0.63)

I'm missing 2 values, how do i solve it from here on?
It helps to define your terms.

What is d supposed to be? What you have written is different from what you wrote in the OP.

Remember, you're supposed to be solving for the angle below the horizontal at which the puck strikes the ground. Is this θ, or something else?
 
  • #9
This is the displacement equation in the horizontal: d=Vi(t)+(1/2)at^2

Since a=0 in the horizontal in cancels (1/2)at^2. I am trying to solve for θ, but this is seemingly the only equation that fits. I feel as though I'm overcomplicating this.
 
  • #10
Im_eNVy said:
This is the displacement equation in the horizontal: d=Vi(t)+(1/2)at^2

Since a=0 in the horizontal in cancels (1/2)at^2. I am trying to solve for θ, but this is seemingly the only equation that fits. I feel as though I'm overcomplicating this.
You can find the horizontal and vertical components of the velocity of the puck as it hits the ground. These should allow you to construct a right triangle using those velocity components as the sides. From there, you should be able to calculate the angle of impact.

Remember, the angle desired is supposed to be measured below the horizontal, so make sure to measure this angle from the correct reference.
 
  • #11
So, it's the angle of impact, not the angle below the table after initial release? Isn't that the angle above the horizontal though? Sorry to be bothersome, but do you mind explaining how to know which angle of reference is to be found by the wording when doing these types of equations?
 
  • #12
Im_eNVy said:
So, it's the angle of impact, not the angle below the table after initial release? Isn't that the angle above the horizontal though? Sorry to be bothersome, but do you mind explaining how to know which angle of reference is to be found by the wording when doing these types of equations?
You've got to read and re-read what the problem asks for, and not jump to any hasty conclusions.

The OP asks this:

"At what angle below the horizontal does the puck hit the ground?"

Hit the ground ≠ below the table after initial release. That was an assumption you made, which unfortunately was in error.

In the second part, about the reference from which the angle is measured, you are assuming that 'the horizontal' is the same as a plane which coincides with the ground. You can measure the angle of impact from the ground up, but the problem here specifically asks for the angle measured below the horizontal. The horizontal can be any plane above and parallel to the ground, like the plane of the table top, for instance.
 

Related to What Angle Below the Horizontal Does a Hockey Puck Hit the Ground?

What is free fall?

Free fall is a type of motion where an object falls towards the ground under the sole influence of gravity. This means that the object is not experiencing any other forces, such as air resistance or propulsion, and is accelerating at a constant rate towards the ground.

What is the acceleration due to gravity?

The acceleration due to gravity is a constant value that represents the rate at which objects accelerate towards the ground under the influence of gravity. On Earth, this value is approximately 9.8 meters per second squared (m/s^2).

How does the mass of an object affect its free fall?

The mass of an object does not affect its free fall. All objects, regardless of their mass, will fall towards the ground at the same rate under the influence of gravity. This is known as the principle of equivalence or the universality of free fall.

What is the difference between free fall and projectile motion?

Free fall is a type of motion where an object is only influenced by gravity, while projectile motion is a combination of both horizontal and vertical motion. In free fall, the object is falling straight down, while in projectile motion, the object is moving both horizontally and vertically.

How can the velocity and position of an object in free fall be calculated?

The velocity and position of an object in free fall can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and time. These equations are v = u + at for velocity and s = ut + 1/2at^2 for position, where v is final velocity, u is initial velocity, a is acceleration, t is time, and s is displacement.

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