A body weighing 2 kg. thrown from the ground vertically upward, fell back. before hitting the ground

A body weighing 2 kg. thrown from the ground vertically upward, fell back. before hitting the ground, it had a kinetic eneology of 20 J. What kinetic energy will the stone have at the top of the pallet trajectory? air resistance to be neglected

Let’s substitute the given value into the formula for kinetic energy:

E kinet = mv² / 2, where E kinet is the kinetic energy of a moving body (unit of measurement – J or kg * m² / s²), m is the mass of the body (unit of measurement is kg), v is the speed of movement of the body (unit is m / from).

We get – 20 J = 2 kg * (v beginning) ² / 2.

From here, we determine the initial speed of our stone –

(v beginning) ² = 2 * 20 J / 2 kg; v start = 4.47 m / s.

A body thrown upwards loses its original speed at every moment of time, because gravity acts against it, which slows down its movement, imparting speed to it in the opposite direction with the acceleration of gravity – v down = g * t (where g is the acceleration of free fall, we will round it up to 10 m / s² for convenience).

The maximum height that a body can rise to is determined by the formula –

h max = v² start / 2g.

In our case, the body will rise to a height h max = (4.47 m / s) ² / (2 * 10 m / s²) = 1 m.

At the top point of its trajectory, at the maximum ascent height, the body stops moving up, “stops”, in order to start free fall with a known downward acceleration.

At this moment, the speed of movement is zero. Consequently, the kinetic energy will also be equal to zero.