Recent content by Dejahboi

Yeah, of course I wouldn't know, that's why I asked for help. Anyways, I have a list of equation I written down from class, but the professor had a tiny mistake categorizing the work of spring and the work energy, and yeah I have ΔK = W Now I lost track of what I'm trying to solve for :/..

W= (1/2)kx2? K = 1/2mx2 Well I got 24.24 m and the book has 25.45 m

For (c) I set -34.45 J = (-0.145 kg) (9.8 N/m) H So, -34.45/ -1.421 = 24.24 m Book says 25.45 m but I can be doing something really wrong.

Homework Statement A 0.145-kg baseball is struck by a bat 1.20 m above the ground, popping straight up at 21.8 m/s. (a) What's the ball's kinetic energy when it leaves the bat? (b) How much work is done by gravity once the ball reaches maximum height? (c) Use your answer in part (b) to find...

Thanks! I'll make note in my notebook with this info. This will help me with my next problem :). THanks again!

Thanks!

Okay so (-2104 J) (145m) = -305080 J?

:(, oh I forgot about part B, alright back to it again

N/kg = J :) My mistake, therefore the kg cancel :)

OMG, thank you! So, 0=21.52+2a(145-0) Set it equal to a, a= -462.25/290 a= -1.59 F=ma F= (1320kg)(-1.59N/kg) =-2104 J

You're right, initial would be 21.5 m/s, and V final would be 0.

W=FΔx

So V= 21.5m/s, V0= 0, X=145, and X0=0 I use F=ma?

Yeah actually saw the typo after I posted the equation lol

So I would use V2=V20+2a(X-X0)?