To find the speed of the water when it hits the bottom, we can use the conservation of mechanical energy principle. Initially, the water has kinetic energy due to its horizontal velocity and gains potential energy as it falls. When it reaches the bottom, this potential energy is converted into kinetic energy.
The initial kinetic energy of the water is given by:
K1 = 0.5 * m * v1^2
where m is the mass of the water (which doesn't matter in this case, since it cancels out) and v1 is the initial horizontal velocity (3.60 m/s).
The potential energy gained by the water as it falls is given by:
U = m * g * h
where g is the acceleration due to gravity (approximately 9.81 m/s^2) and h is the height of the falls (108 m).
At the bottom, the total kinetic energy of the water is given by:
K2 = 0.5 * m * v2^2
where v2 is the final speed of the water.
Using conservation of mechanical energy (K1 + U = K2), we can find the final speed of the water:
0.5 * m * v1^2 + m * g * h = 0.5 * m * v2^2
Notice that the mass (m) cancels out:
0.5 * v1^2 + g * h = 0.5 * v2^2
Now plug in the values and solve for v2:
0.5 * (3.60)^2 + 9.81 * 108 = 0.5 * v2^2
6.48 + 1058.68 = 0.5 * v2^2
1065.16 = 0.5 * v2^2
v2^2 = 2130.32
v2 = √2130.32 ≈ 46.15 m/s
So, the speed of the water when it hits the bottom is approximately 46.15 m/s.
the great gray-green, greasy zambezi river flowa over victoria falls in south central africa. the falls are approximately 108 m high. if the river is flowing horizontally at 3.60 m/s just before going over the falls, what is the speed of the water when it hits the bottom? assume the water is in freefall as it drop.
1 answer
1 answer