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So for this equation a=1 (1=1a, 1/1=a, 1=a). What equation(s) would I use after this?

I used an online simulation for this lab. Mass: 1.0 kg Force: 1 N X position: – 2 Y position: 0 The final position was 6.0, final velocity was 4.0, and final time was 4.0 At 3 seconds the position was at 2.50 and velocity was at 3.00 Because I used an online simulation I...

The angle of the cannon is 50°.

I was given these equations and was told to solve for V: t = 12/Vx Vy = 6/t + (0.5)(9.81)t Vx = Vcos(50) Vy = Vsin(50) And this is what I got: t = 12/Vx 2.6=12/Vx Vx=28.8 Vy = 6/t + (0.5)(9.81)t Vy=6/2.6+(0.5)(9.81)(2.6) Vy=14.272 Vx = Vcos(50) 28.8/cos(50)=V V=29.8456 Vy = Vsin(50)...

I used an online simulation for this lab. The target is 12 m away horizontally and 6 m vertically. I was to find the velocity needed to hit the target.Since it's an online simulation, I played around with it and got 14.3 m/s. The formula I plan on using is: Vx=range/time...

Thanks! This made is a bit more clear to me. I found got this equation, and your description made it make sense: Vx=16cos(∡)=26.7m/2.6s Vy=√(256-(26.7m/2.6s)²) V=16=√(26.7m/2.6s)²+Vy²

For the most part, yes. I used an online simulation for this lab and just continually changed the value of the initial velocity until it hit the target.

To be honest, I'm not too sure. My instructions state, "Calculate the theoretical initial velocity (no air resistance) required to impact a target with the distance you used in the simulation." I actually messed up when typing the original post: range quantity is the same as target (25.6 m).

I can do with just an equation, but if anyone is willing to help: Target is 25.6 m away from canon Initial Velocity = 16 m/s Time = 2.6 s Range = 26.7m No air resistance

I can do with just an equation, but if anyone is willing to help: Target is 25.6 m away from canon Initial Velocity = 16 m/s Time = 2.6 s Range = 26.7m No air resistance