Physics about motion

At the beginning,
Kinetic energy of the stone = ½ mu², where u is the initial velocity = 20 m/s

Just before hitting the ground,
energy possessed = kinetic energy at that time
= Initial kinetic energy + loss in potential energy due to falling
= ½ mu² + mgh
= ½ (2)(20)² + 2*10*15
= 400 + 300
= 700 J

So the energy possessed by the stone just before touching the ground is 700 J.


2007-01-06 23:43:20 補充：
In this case, you don't need to care about the resultant force.If it is required to use the acceleration such as to find the trajectory the stone is falling or velocity at any time, we need to resolve the vectors into horizontal and vertical components. This leaves to the A-Level courses.

2007-01-06 23:51:53 補充：
Note that during the fall, the weight of the stone is its only resultant force.The calculation in the vertical direction is the same as free falling objects. Only the equation of motion along the horizontal direction is different. The trajectory forms a parabola.

For moment just leave the cliff
the 2kg stone have kinetic energy and potential energy

Total energy = 0.5mv^2 + mgh
i.e. =0.5 x 2 x 400 + 2 x 10 x 15 = 400 + 300 =700J

Base on energy conservation, no energy loss from cliff to just before touching the ground
i.e. 700J at all the time!!

We can consider the stone dropped from height, stone will loss potential energy due to H(height) decrease...

Energy conservation again, the potential energy will change into kinetic energy and lead the stone to fly more faster to ground than beginning...