Using conservation of energy, find the maximum height Hmax to which the object will rise.

**Answer 1**

CONSERVATION OF ENERGY:

kinetic energy gained

=

potential energy lost

and vice-versa

So if you start with no PE and end with no KE, that becomes

Initial KE = 1/2 mv^2

=

final PE = mgh

Solve for final max height starting from a speed v:

h = v^2 / 2g

Or going the other way, solve for speed after a fall from height h:

v = sqrt (2gh)

**Answer 2**

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

**Answer 3**

Vb=sqrt(v^2+2gh)

Source(s): I just did it

**Answer 4**

Hmax=((V^2)/(2g))

**Answer 5**

h=v^2/2g

**Answer 6**

I believe I actually have to agree with ashhh on this one

**Answer 7**

what the frick?