Maximum height using conservation of energy?

Lizzette Wendroth: I believe I actually have to agree with ashhh on this one :(

Jeannine Vassie: Vb=sqrt(v^2+2gh)

Russel Gajate: h=v^2/2g

Nona Lentini: what the frick?

Darnell Cutliff: Hmax=((V^2)/(2g))

Corrinne Ruozzo: CONSERVATION OF ENERGY:kinetic energy gained=potential energy lostand vice-versaSo if you start with no PE and end with no KE, that becomesInitial KE = 1/2 mv^2=final PE = mghSolve for final max height starting from a speed v:h = v^2 / 2gOr going the other way, solve for speed after a fall from height h:v = sqrt (2gh)...Show more

Mauro Cowee: A greater finished answer usinfc3367785ba1c1ae27ef4bdd2cb1d05b enerfc3367785ba1c1ae27ef4bdd2cb1d05by is PE(i) + KE(i) + W(nc) = PE(f) + KE(f) 0 + ½ m fc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05b + 0 = m v²/2g h + ½ m (v²/2g cos ?)v²/2g the place v²/2g cos ? is the horizontal element of the ball's fc3367785ba1c1ae27ef4bdd2cb1d05belocity, which does no longer! chanfc3367785ba1c1ae27ef4bdd2cb1d05be. h = [fc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05b] (a million - cosfc3367785ba1c1ae27ef4bdd2cb1d05b?) observe that if ? = ninety°, you fc3367785ba1c1ae27ef4bdd2cb1d05bet h = fc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05bfc3367785ba1c1ae27ef4bdd2cb1d05b as predicted...Show more