# Flight path of a disc

1. Jul 17, 2013

### Zondrina

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

$θ = 28°$
$\vec{Δd}_H = 39.8 m [F]$

3. 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.

Last edited: Jul 17, 2013
2. Jul 17, 2013

### PeterO

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. Jul 17, 2013

### Zondrina

Yeah I guess disc can be interpreted as a discus. I don't think the question is intending it to be a frisby.

4. Jul 17, 2013

### PeterO

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. Jul 17, 2013

### Zondrina

$cos(28°) = 0.88$ and then (0.88)(2) = 1.76.

6. Jul 17, 2013

### PeterO

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. Jul 17, 2013

### haruspex

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. Jul 17, 2013

### Zondrina

Yeah I definitely should use that identity more often. It makes it easier to look at for sure. Thanks for your help :).

9. Jul 17, 2013

### PeterO

True: your adoption of the 0.88 value loses the fact it is really 0.882964 so a small variation comes in.

10. Jul 17, 2013

### Zondrina

I don't really mind the inaccuracies of the computations too much, it's the understanding I care for.

11. Jul 17, 2013

### haruspex

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.