Someone throws a 1k mass vertically on a balcony 10m above the ground at a speed of 10ms

Analysis: Because the work that people do to the ball is only at the moment before the ball is thrown, when the person is in the air, the person does not do work on the ball (your hands can't fly with the ball, right?). So it can be based on:

W = 1 2mv Find the work done by the person on the ball. At the height of H above the ground, the kinetic energy of the grapefruit is equal to one-half of the gravitational potential energy, and the velocity is vt, then there is mgH=mvt, because the grapefruit is only subjected to gravity during motion, so it can be found by the conservation of energy: mgh+1 2mvo =0+1 2mvt (h=10m, m=1Kg).

Solution: W = 1 2mv

So W = 1 2 1Kg 100m s

50J is set to be H above the ground and the instantaneous velocity is vt1

There is mgH=mvt1, and both sides go about m

H=vt1²/g

vt1 -vo =-2g(H-10m) vt1 =vo -2g(H-10m).

So H=vo g-2 (H-10m).

30m=3H

H=10mLet the ground be 0 force surface, so the gravitational potential energy when it reaches the ground is 0, so mgh+1 2mvo =0+1 2mvt (h=10m, m=1Kg).

100J+50J=vt²/2

So vt=10 root number under 3m s

Solution: 1. People do work on objects to make grapefruit gain kinetic energy.

W=1 2mV2=1 2 1 (10)2=50J2, the total kinetic energy of the grapefruit in the air comes from the conversion of the work done by the beginning man to the grapefruit 50J and the gravitational potential energy of the grapefruit falling.

1 2 mV2 + mg (10-h) = mgh where 1 2 mV2 = 50J (first asked).

The solution is h m3, the conservation of mechanical energy (all gravitational potential energy is converted into kinetic energy when landing, and the ground is zero potential surface) 1 2mV2+mg 10=1 2mVo2

Solve Vo=300 open root number.

What to ask? It landed on the ground about left and right.

1)W=Ep=1/2mv^2=50J

2) E total mechanical energy = W + mgh = 150J

E potential = mgh = 100J

h=10m