Divers in Acapulco, Mexico, leap from a point 36m above the sea. What is the velocity when they enter the water? V = distance / time

We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

Initial velocity, u = 0 m/s, Displacement = 36 m , acceleration = acceleration due to gravity = g = 9.8 [tex]m/s^2[/tex]. We need to find final velocity.

Substituting

[tex]v^2=0^2+2*9.81*36\\ \\ v=26.58 m/s[/tex]

Velocity Divers in Acapulco, Mexico when they enter the water from a height of 36m above the sea = 26.58 m/s

facundo3141592 facundo3141592 · Answer 2 · 2020-03-16T01:41:30+01:00

Answer: The velocity is 27.6m/s

Explanation: We know that they leap from a distance of 36m.

We know then that the initial height is equal to 36m, and this means that the initial potential energy is equal to:

U = mgh

where m is the mass, g is the gravity acceleration and h is the height.

Right in the point where the diver touches the water, we have h = 0, and this means that all the potential energy transformed into kinetic energy, that is written as:

K = (1/2)*m*v^2

And from here we can find the velocity in that point.

m*g*h = (1/2)*m*v^2

2*g*h = v^2

v = √(2*g*h)

and g = 9.8m/s^2, h = 39m

v = √(2*9.8m/s^2*39m) = 27.6m/s