Collision

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NA19

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Two guns, separated by five meters and pointed at each
other, are tilted so that each gun points at an angle  above
the horizontal. Both guns are fired simultaneously, with
each bullet leaving its gun at a speed of 250 m/s. The bullets
collide inelastically in midair, falling together to the
ground.

How much time elapses between the firing
of the guns and the collision of the two bullets?

I tried to use the equation vy = v0y - gt
So I got 25sintheta = t

However, the answer is apparently 0.01/costheta.

If someone could explain, that would be great!
 
They're 5 meters apart, shoots at the same velocity. The bullet will collide at the midpoint. (2.5 meters away from each gun) You can say that the horizontal displacement is 2.5 meter, and x=V0*cos(theta)t. This gives t=x/(V0*cos(theta))=2.5/(250*cos(theta))=0.01/cos(theta)
 
Not dy, vy = v0y - gt. Why is 25 sintheta wrong?

NA19 said:
I tried to use the equation vy = v0y - gt
So I got 25sintheta = t

First you're using an approach that doesn't make sense. To find the time at which they collide, you can't use vy. Vy is always the same for both bullets. The only way to find the point of collision in this case is by setting Dx of one bullet=Dx of another.

Second your algebra is wrong. Vy=V0y - gt. V0y is definitely 250 sin(theta), but you obviously ignored Vy and simply set V0y/g=t
 
yeah you want the horizontal velocity, V*cos theta
then, since v = meters / time, you can do meters divided by the horizontal velocity
2.5/ 250 cos theta
0.01 / cos theta
 
Seems like this scenario could result in vy=0 upon collision if given the proper value for theta, but doesn't have to
 
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