Recent content by Hadjiev

Dan, I want to thank you. You found my mistake. If you drop by this post could you explain my mistake in a bit more detail. My email is Nick_Hadjiev@hotmail.com. I would just like to clear it up incase I encounter a problem like this in the future. Thanks again, Note: nevermind, my friend...

It's good that everyone is grasping the concept. I forgot to put that the distance from the sniper to the target is 1500 m, which is not the entire distance travelled, but merely the distance from the sniper to the target. [SIZE="6"]The distance from the sniper to the target is 1500 m Ok...

I think velocity has to be negative in this situation. The bullet travels a certain distance, reaches maximum height, and then comes down to the target. Thus, the vertical component is doing something like this: *| || <---- 40 m |* | | Haha, bear with the pathetic diagram...

the 1.756 came from this: Horizontal: v = d/t 854cosx = 1500/t t = 1500/854cosx t = 1.756/cosx Since we know the time of flight for the horizontal is the same time it takes for the bullet to travel up to its maximum and then down to the target, we can substitute time into the vertical...

Here goes: A sniper shot is fired at over a kilometer away. Furthermore, the target is at the very edge of a building that is 40 m high. Solve for the elevation angle that will be required to score a perfect hit. The initial velocity of the bullet is 854 m/s. Note: the bullet is not at max...

I am so confused. I'm going to skip it and hope that a question like that isn't on my test. Thanks for your time.

I did just that. I found the midpoint of each side, took the slope, and then determined the equation of each right bisector. My one equation is y=a/2, whereas my other two eqns are huge. Is there a way to simplify the other two equations using y = a/2? Can you please show me how to prove this...

The question asks: In triangle ABC, with vertices A(0,a), B(0,0), and C(b,c), prove that the right bisecors of the sides meet at a common point. Ok, this question is really getting to me. I know that I could find the equations of all three right bisectors and then solve for x and y through...

Ack! I forgot about the upward motion. Thanks Astronuc.

Hey, I was wondering if someone could check my answer for this question: A helicopter is rescuing a boy from a flooded street. Hovering 20 m in the air, a cable is lowered to the child. After it is attached, the helicopter begins to accelerate upwards at 0.2 m/s^2. Eight seconds later, the...