A really though question

1. Oct 25, 2005

404

A really tough question

well.. for me atleast
Any hints? I'm still trying to figure out where to start

EDIT: *though = tough
...can someone please tell me what the answer would be? So atleast I would know if I solved it right or not.

Last edited: Oct 25, 2005
2. Oct 25, 2005

lostdaytomorrow

Both of their y-components are equal... The x - component of the bullet is just there to get in the way. Assuming no friction or air resistance

3. Oct 25, 2005

saschouch

well, you know that the acceleration of gravity is -9.81 m/s^2 right? that has something to do with it. does it say if they start at the same height?

4. Oct 25, 2005

404

no, the gun is below the target, but it is aimed straight at the target.

Well, I suppose their time value would be the same as well.

5. Oct 25, 2005

saschouch

well, you know that the acceleration of gravity has no effect on the horizontal component, and no matter the mass of the object, the downward acceleration will be the same for both. the velocities will be different, because of mass differences, but you can still use that information to solve it.

6. Oct 25, 2005

404

Yeah, I already know all the things talked about above from all the other projectile motion questions I did... I made some progress, but can someone please tell me what the answer would be? So atleast I would know if I solved it right or not...

7. Oct 25, 2005

saschouch

well, unless you have specific numbers for like the velocity of the bullet, or masses, then there's no number-plugging work, obviously. this link seems to show it accurately: http://www.physicsclassroom.com/mmedia/vectors/mzf.html
if that doesn't help you, then i dunno what to do, because monkeys and bananas can solve ALL problems.

8. Oct 25, 2005

404

lol... thanks, I read it, but it doesn't seem to give an algebraic solution.

9. Oct 25, 2005

404

bump, does anyone know?

10. Oct 26, 2005

daniel_i_l

Lets say that the distance between them on the x- axis is X, and on the y-axis it is Y, the target's hight is H. The components of the projectile's initial speed are Vx and Vy, without gravity they should hit with those speeds. So the time that it takes the projectile to get to the target is X/Vx we'll call that time T. T is constant with or without gravity (as they mentioned above). With gravity we can write that the projectile's hight is:
h = Vy*t - 1/2*gt^2 and the target's hight is:
h = H - 1/2*gt^2
we know that for them to collide they have to have the same hight at time T (thats when they have the same x). we also know that Vy*T = H because without gravity they would have hit at hight H (the target woulden't have fallen, so after time T the projectile's hight is:
hp = H - 1/2*gT^2 and the target's hight is:
ht = H - 1/2*gT^2 so:
ht = hp!

11. Oct 26, 2005

404

Oh I see now. Thanks