What is the Velocity of a Disc Thrown at a 28° Angle to the Ground?

In summary: If it does, then I clearly have a problem with my understanding at that point.OK, but I would still urge working symbolically until the last. It makes it much easier for others to follow the logic and easier for everyone to spot mistakes. A trick I often use after completing the algebra is to pick a set of easy values, substitute them at various points in it, and see if the answer it gives suddenly changes. If it does, then I clearly have a problem with my understanding at that point.In summary, the disc, resembling a discus, is thrown at a 28° angle to the ground and lands 39.8m away. Using the equation ##\vec{Δd}_H =
  • #1
STEMucator
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Homework Statement



A disc is thrown at a 28° angle to the ground. It landed 39.8m away.

a) What is the velocity of the disc?
b) How long was the disc in the air?

Homework Equations



##θ = 28°##
##\vec{Δd}_H = 39.8 m [F]##

The Attempt at a Solution



a) I was thinking I should use the equation ##\vec{Δd}_H = \vec{v}_H Δt##. I of course don't have a time or horizontal velocity to work with, but what I do have is that :

##\vec{v}_H = \vec{v}_R cos(θ)## and subbing back I get ##\vec{Δd}_H = \vec{v}_R cos(θ) Δt## ( Note vR is the resultant velocity ).

Now as for ##Δt##, I want to replace it with ##\frac{2 \vec{v}_R sin(θ)}{\vec{a}}## as I can work with it.

So plugging back into my equation a second time and moving things around a bit I get :

##\vec{v}_{R}^{2} = \frac{ (39.8)(9.8) }{ (1.76)(sin(28°)) }##
##∴ \vec{v}_{R} = 22 m/s##

b) I simply use ##Δt = \frac{2 \vec{v}_R sin(θ)}{\vec{a}}## and I get ##Δt = 2.1 s##.

Part a) is the one I'm concerned with. Hopefully it looks okay.
 
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  • #2
Zondrina said:

Homework Statement



A disc is thrown at a 28° angle to the ground. It landed 39.8m away.

a) What is the velocity of the disc?
b) How long was the disc in the air?

Homework Equations



##θ = 28°##
##\vec{Δd}_H = 39.8 m [F]##

The Attempt at a Solution



a) I was thinking I should use the equation ##\vec{Δd}_H = \vec{v}_H Δt##. I of course don't have a time or horizontal velocity to work with, but what I do have is that :

##\vec{v}_H = \vec{v}_R cos(θ)## and subbing back I get ##\vec{Δd}_H = \vec{v}_R cos(θ) Δt## ( Note vR is the resultant velocity ).

Now as for ##Δt##, I want to replace it with ##\frac{2 \vec{v}_R sin(θ)}{\vec{a}}## as I can work with it.

So plugging back into my equation a second time and moving things around a bit I get :

##\vec{v}_{R}^{2} = \frac{ (39.8)(9.8) }{ (1.76)(sin(28°)) }##
##∴ \vec{v}_{R} = 22 m/s##

b) I simply use ##Δt = \frac{2 \vec{v}_R sin(θ)}{\vec{a}}## and I get ##Δt = 2.1 s##.

Part a) is the one I'm concerned with. Hopefully it looks okay.

Without reading your solution, there is a simple question.

Are you to consider this disc as a simple projectile, or is the flight more like a Frisby? If it is like a Frisby, you need much more information.

Are you actually talking about a discus?
 
  • #3
PeterO said:
Without reading your solution, there is a simple question.

Are you to consider this disc as a simple projectile, or is the flight more like a Frisby? If it is like a Frisby, you need much more information.

Are you actually talking about a discus?

Yeah I guess disc can be interpreted as a discus. I don't think the question is intending it to be a frisby.
 
  • #4
Zondrina said:
Yeah I guess disc can be interpreted as a discus. I don't think the question is intending it to be a frisby.

If this amounts to a simple projectile, analysis of the vertical and horizontal components leads to a simple range formula, which you may find in your text or on the web.

That formula ultimately involves using the sine of 56o rather than 28o, so I am not sure of your calculations.

where did your 1.76 value come from?
 
  • #5
PeterO said:
If this amounts to a simple projectile, analysis of the vertical and horizontal components leads to a simple range formula, which you may find in your text or on the web.

That formula ultimately involves using the sine of 56o rather than 28o, so I am not sure of your calculations.

where did your 1.76 value come from?

##cos(28°) = 0.88## and then (0.88)(2) = 1.76.
 
  • #6
Zondrina said:
##cos(28°) = 0.88## and then (0.88)(2) = 1.76.

OK, in that case your figures should be correct.

If you had left your denominator uncalculated it would have read 2.cos28°.sin28°, which, using the double angle formula (sin(2A) = 2.sinA.CosA) equals the sin56° I was expecting to see.

The range formula simplifies to Range = [V2.sin(2θ)]/g or (V2/g).sin(2θ)

which shows maximum range is when sin(2θ) = 1 or θ = 45°.
 
  • #7
Zondrina said:
##∴ \vec{v}_{R} = 22 m/s##
Seems a little inaccurate. To preserve accuracy, it's best to work entirely symbolically until the final step. What expression do you get for vR in terms of g, θ and dH if you do that?
 
  • #8
PeterO said:
OK, in that case your figures should be correct.

If you had left your denominator uncalculated it would have read 2.cos28°.sin28°, which, using the double angle formula (sin(2A) = 2.sinA.CosA) equals the sin56° I was expecting to see.

The range formula simplifies to Range = [V2.sin(2θ)]/g or (V2/g).sin(2θ)

which shows maximum range is when sin(2θ) = 1 or θ = 45°.

Yeah I definitely should use that identity more often. It makes it easier to look at for sure. Thanks for your help :).
 
  • #9
haruspex said:
Seems a little inaccurate. To preserve accuracy, it's best to work entirely symbolically until the final step. What expression do you get for vR in terms of g, θ and dH if you do that?

True: your adoption of the 0.88 value loses the fact it is really 0.882964 so a small variation comes in.
 
  • #10
I don't really mind the inaccuracies of the computations too much, it's the understanding I care for.
 
  • #11
Zondrina said:
I don't really mind the inaccuracies of the computations too much, it's the understanding I care for.
OK, but I would still urge working symbolically until the last. It makes it much easier for others to follow the logic and easier for everyone to spot mistakes. A trick I often use after completing the algebra is to pick a set of easy values, substitute them at various points in it, and see if the answer it gives suddenly changes.
 

1. What factors affect the flight path of a disc?

The flight path of a disc is affected by a variety of factors, including the speed and direction of the throw, the angle at which the disc is released, the wind conditions, and the shape and weight of the disc itself.

2. What is the difference between understable and overstable discs?

An understable disc is one that tends to curve to the right (for a right-handed thrower) when thrown with the same amount of power and angle, while an overstable disc will curve to the left. This is due to differences in the disc's shape and weight distribution.

3. How does the spin or rotation of the disc affect its flight path?

The spin or rotation of a disc is what gives it stability and helps it maintain its desired flight path. A disc with more spin will have a more stable flight and will be less affected by wind or other external factors.

4. Can the flight path of a disc be affected by the temperature?

Yes, the temperature can affect the flight path of a disc. Colder temperatures can make the disc less stable and cause it to fly less predictably, while warmer temperatures can make the disc more stable and result in longer flights.

5. How can I improve my understanding of the flight path of a disc?

The best way to improve your understanding of the flight path of a disc is to practice and experiment with different throws, discs, and conditions. You can also watch tutorials and videos from experienced disc golfers, and try to understand the physics behind disc flight.

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