The height of a certain waterfall is 33.2 m.?

Question

When the water reaches the bottom of the falls, its speed is 25.9 m/s. Neglecting air resistance, what is the speed of the water at the top of the falls?

Answer 1

let the speed of the water on top of the waterfall be v and the speed at the bottom be Vf,

so,

Kinetic energy of water at top + Gravitational portential energy at top = Kinetic energy at bottom

1/2 m v^2 + mgh = 1/2 mVf^2
1/2 v^2 = 1/2 Vf^2 - gh..................dividing both sides my a constant m and rearranging

v^2= Vf^2 - 2gh
v = root( (25.9)^2 - 2(9.81)(33.2) )
= 4.407 m/s

Answer 2

The fall adds kinetic energy which will increase the speed by
sqrt(2*g*h)

or
v=sqrt(v1^2+2*g*h
v1=sqrt(25.9^2-2*9.81*33.2)
v1=4.4 m/s

j

Answer 3

vve want the v0 the first speed
use this formula
v^2-v0^2=-2gy
(25.9)^2 - v0^2 = -20 x -32.2
from here you can find the v0

the reason that we had -32.2 is becuse it is downward