Could someone check my solution to this?

A shell explodes on the surface of horizontal ground. Earth is scattered in all directions with
varying velocities. Show that particles of earth with initial speed v landing a distance r from
the centre of explosion will do so at times t given by
$$\frac{1}{2}g^2t^2=v^2 \pm \sqrt{v^4-r^2g^2}$$

Question

Could someone check my solution to this?

A shell explodes on the surface of horizontal ground. Earth is scattered in all directions with
varying velocities. Show that particles of earth with initial speed v landing a distance r from
the centre of explosion will do so at times t given by
$$\frac{1}{2}g^2t^2=v^2 \pm \sqrt{v^4-r^2g^2}$$

anonymous · Answer

http://gyazo.com/50f04a7f09df3fa4192bceae0367fb74

anonymous · Answer

I half-reached the solution, but the plus-minus killed my method, as in general $$\sin^2x \ne \cos^2x$$

Where did I make the logical error?

zehanz · Answer

But will the particles not also have all kinds of vertical angles $\theta$ at which they are being scattered? Then $v \sin \theta$ would determine the time needed to get to the ground again.

anonymous · Answer

Of course, which is why I have left the angle as arbitrary. However, that makes little inroads into the question.

zehanz · Answer

I've got to the part that says $v^4-g^2......$. Where did that come from?

anonymous · Answer

That's \[v^4-g^2r^2\[, the thing inside the square root

anonymous · Answer

$$v^4-g^2r^2$$

anonymous · Answer

Substituting in what I found for r.

zehanz · Answer

But inside the square root in your question it was $v^2$...

anonymous · Answer

That was a typo- sorry!!

anonymous · Answer

http://gyazo.com/c7996dccd0b5b0a9d95d7161217a4872 Here's the original question as I'm evidently incompetent to duplicate it correctly.

zehanz · Answer

OK, I'll go on with it...

anonymous · Answer

Thanks

zehanz · Answer

I get everything the same as you:$$\sqrt{v^4-g^2r^2}=\sqrt{v^4\cos^22w}=v^2|\cos2w|=\pm v^2\cos2w$$So$$v^2 \pm v^2\cos2w=v^2(1 \pm \cos2w)=v^2(1 \pm(1-2\sin^2w))$$The plus or minus leads to $2v^2\cos^2w$ and $2v^2\sin^2w$.

I do not see any errors in your calculation.
Could it have something to do with the value of w?
If w > 45 deg, then r becomes smaller again, and cos2w (in the first line of this reply) becomes negative.

anonymous · Answer

The problem is that this is supposed to be equal to $$0.5g^2t^2$$
Which is $$2v^2\sin^2(w)$$
However, one of those solutions was$$2v^2\cos^2(w)$$, which is obviously not the same.

zehanz · Answer

Yeah, that was my problem too...