2 Projectiles

Hey There. Am Hitting a bit of a brick wall with this question.

So, initially I thought It would be fairly simple. I made 2 equations for the Y displacement and 2 equations for the x displacement (Defining the left hand projectile as the origin, so the initial displacement of the right hand one would be equal to d).

I then defined the time for the right hand projectile to be t-0.5, as its fired later than the other one. From here I was able to equate both the equations for Y displacement to get a value of t=1.37s. Seemed a bit small to me initially, but I continued.

I then evaluated the 2 X displacement equations, now having the time at which the collision occured, allowing me to have the initial displacment of the 2nd projectile, and hence, d. Giving me a value of d=108.1m.

However, the answer given for the questions is 302.1m. It's hard for me to see where I went wrong as my method managed to go all the way to an answer, but it just isn't correct. Perhaps my method is right but I've made some error on the way? Not sure. Will post a picture of my working if sometone wants but its a tad messy.

So Is what I did fundamentaly right? Or have I neglected something?

Cheers,
Eremor.

collisions.jpg121.1KB

projectiles.jpg52.6KB

Reply 1

Stonebridge

My first thought is that if you define t=0 as the time the left projectile is launched, the second one, on the right, is launched at t + 0.5s if it starts 0.5s later.
Does that make any difference to your answer?

Reply 2

Eremor

Might be doing something wrong here but when doing that, I actually come out with a negative answer for some strange reason.

Just considering the Y Direction, I come out with the equation

V1t - 0.5gt^2 = V2(t+0.5)-0.5g(t+0.5)^2

Expanding and then re-arranging for t gave me

t= (0.5V2-(1/8)g)/(V1-V2+0.5g)

Coming out as -1.52s (As V2 is much larger than V1 making the denominator negative). I think It must be something wrong with my working, so will go back over it all again, as I cant think of another way to do it.

Sorry if my working is a bit tricky to understand, but its just how I wrote it. Projectile 1 is on the left, 2 on the right. V= initial velocity

(edited 9 years ago)

Reply 3

Stonebridge

Regarding the time, although what I said is correct re the right object starting later at t+0.5, this does mean that the left object is in flight for 0.5s longer than the right. So in the equations of motion, the time for the left object is t+0.5 if the right is t. Sorry if I confused you.

First consider the horizontal motion
If obj 1 travels, say, x and obj 2 a distance y then of course x+y=d, the value we want.

1)
x = (200 Cos 45)(t +0.5)
2)
y = (250 Cos 60)(t)

To find t use the vertical motion and use
s = ut + ½gt²

Both objects travel the same vertical height h if they collide. Assuming they collide on the way up*
1)
h = (200 Sin 45)(t+0.5) + ½g(t+0.5)²
2)
h = (250 Sin 60)(t) + ½gt²

Equate 1 and 2 to get t

Sub t into the horizontal equations to get x and y

*One thing about this question is that there can't be a single unique value for d, because it is possible, I imagine, for the objects to collide
a. when both rising
b when both falling
c. when one is rising and the other falling

Just a thought.

(edited 9 years ago)