How do you calculate the kinetic energy of a falling object?

When the rock is 1.90 m above the water, it is already moving at 6.10m/s.

So it's kinetic energy is initially ½mv² = ½ x 68.0 x 6.10² = 1265J

It falls a distance 1.90m so potential energy lost = mgΔh = 68.0 x 9.8 x 1.90 = 1266 J

The kinetic energy gained equals the potential energy lost.

So the total kinetic energy when it hits the water = 1266+1267 = 2533J = 2530J to 3 significant figures.

Oh boy are you making this hard.

What's the definition of KE? You really should learn that. You'll be using it over and over in physics.

Anyway KE = 1/2 mV^2 by derivation from the basic physics, and that V^2 = U^2 + 2gH where U = 6.10 mps at H = 1.9 m. g is g = 9.8 m/s^2. Plug and chug.

What you and the other answers ignored with your mgh is that the rock already had some KE at h = 1.9 m because it already had some speed 6.1 mps.

a million) the fee as a function of time is merely V(t) = g t for this reason the kinetic power is KE = a million/2 m V^2 = a million/2 m g^2 t^2 use your calculator to discover the ability 2) the full power of the ball is 0, even nonetheless, the full power is the sum of skill power and kinetic power. for this reason, after falling 1m, the skill power equals the kinetic power, it fairly is m g h = a million/2 m V^2 for this reason, the kinetic power is merely KE = m g h = (a million) (9.8) (a million) = 9.8 J

PE = mgh or wh (w being weight, weight = mass x acceleration due to gravity)

KE = 1/2mv^2

PE at the top is the same as the KE at the bottom.

PE looks for distance, while KE looks for if the object is moving.

If the object isn't moving, it has no KE.

Here's how I think it would be done and I believe I'm right. (check your book and tell me what the answer is if I'm wrong and I'll try to rework it)

KE = 1/2mv^2

KE = 1/2 (68) (6.10)^2

KE = 34(37.21)

KE = 1265.14 J (Joules)