Projectiles

Not sure where the other replies to this thread have gone. But they don't collide. Something is wrong/missing in the question.

She's reposted it. Really annoying when people do that...

Yep. Just noticed it. She can repost it as often as she likes but it won't change the fact that they won't collide for exactly the same reason that if you throw a banana horizontally off the top of a cliff and someone simultaneously fires a rifle at the bottom, then the banana will remain unscathed unless the angle of elevation of that bullet is somewhat steep

Using A, resolve horizontally using SUVAT to find the time when A has travelled 40 meters horizontally. You then need to resolve vertically (gravity!) to find the vertical displacement

Then you resolve B both horizontally and vertically (using the time from A) to prove the displacement vectors are the same

Hope this helps!

I'm sorry. The re-post was genuinely unintentional. I was working on an iPad and I thought I hadn't sent it at all.

Here is the question in the attachment

Thanks. What textook is it (in case I have it lying about), or please send a piccie of the solution in case we're missing something.

Not sure where the other replies to this thread have gone. But they don't collide. Something is wrong/missing in the question.

Its a bit hard to read your images (answer and question), but I'm presuming they're using the (standard) equation of a projectile
y=xtanθ - gx2/2u2cos2θ
l've not verified it, but I have little doubt they do both pass through that point. But I'd suspect that its at different times so they do not collide.

There is no time in the formula, its simply a quadratic in x. Using standard SUVAT will involve time as you're integrating up acceleration, hence its doomed to failure to find a point of intersection as they don't pass through the same point at the same time, hence no collision.

When you upload an image, could you check its not too blurred etc. It makes it easier for people to help. Thanks.

Here is the question word for word from the text book: (apologies for blurred photos)

Two particles, A and B, are launched at the same time in the same vertical plane, with A 60 m vertically above B. Particle A is launched horizontally and particle B at 45 degrees above the horizontal. The particles are launched at speeds 14 m/s and 28 m/s respectively. Show that the particles collide after travelling 40 m horizontally.

It definitely uses the word collide.

The answer in the book uses the equation y= xtanT - (gx^2)(1 + tan^2 T)/(2U^2) where T is the angle of projection above the horizontal and U is the initial speed.

I've never seen this equation before and was trying to use suvat. If they collide then the length of time they are in the air must be the same and their vertical distances added together must equal 60. But it didn't work out and I can't figure out why. If you have any explanation for that I would be most please to hear of it.

Thanks.

We all know that. Trouble is that it doesn't work in this situation as the question is ill conceived.

The question indicates that t=40/14 for A
Trouble is that for B, the horizontal distance travelled is thus 40 / 14 root 2

So, they do not meet

I’m afraid you’re wrong and it does indeed work with suvat. See the attached image;

The Times for each particle are different. They are not in the same place at the same time. They do not collide.

Yes the times for each particle are different, this is because they have different speeds of projection so they MUST be different. Think about it you can have 2 people running 100m, the one who runs faster gets to 100m first.

We agree about them passing through the same point and at different times. Do you agree about the fact that they don't collide?

I agree that the question is poorly phrased and if you conduct the question properly they shouldn’t collide, but I believe this text book will be looking for the answer 40 20