Recent content by xasn23

I got both of them already. The first one I have to use Acceleration= [(Final Velocitysquared - Initial Velocitysquared)]/ 2*distance in m so I changed 2.3 cm > .023m And then 0-25^2 625m/s A= _______ = _____ = 13,586.957 m/s 2 (.023m) .046m For the second one. All I had to find...

oh yeah. I got both of them already. The first one I have to use Acceleration= [(Final Velocitysquared - Initial Velocitysquared)]/ 2*distance in m so I changed 2.3 cm > .023m And then 0-25^2 625m/s A= _______ = _____ = 13,586.957 m/s 2 (.023m)...

my textbook doesn't help me because it uses different situations... but it doesn't help me understand the 45 degrees part... do I find the acceleration first like so: (sin45*1620) or something of that sort... I'm still not sure how to. I'd really appreciate it if you helped me out...

no... I don't ... I just learned this stuff a few hours ago.

... dangit ... Wait ok so since I'm dealing with a 45 degree drop... does that mean I'll take the F=MA and divide it by 2? I'm pretty sure that that's what I'm supposed to do...

ahh ok I see thank you! >:D<

I'm not sure what the final and initial velocity is ... I'm thinking that the final is 0 since the axe stops in the end... But I'm still skeptical about the initial... how do I find the initial? Please help? =)

[SOLVED] Homework: Forces and Acceleration 1.) While chopping down his father's cherry tree, Anthony discovered that if he swung the axe with a speed of 25 m/s, it would embed itself 2.3 cm into the tree before coming to a stop. a.) If the axe head had a mass of 2.5 kg, how much force was the...