a boy standing on top of a building throws a stone to hit an object on the ground. state the energy changes that occur in this event.
4 answers
I don't even know it, that's why I asked the question, pls if u know it pls tell me. thanks.
Lacking significant data. Need to know height of building. Also, if he "throws" it, need to know power of his throw.
The potential energy goes down by an amount m g h where m is ball mass, g is gravity acceleration and h is the height of the pitchers hand above ground.
The pitch is thrown with horizontal speed u and vertical speed v.
The kinetic energy due to the horizontal speed (1/2) m u^2 remains constant until ground except for air friction.
However v is tricky because it might be initially upward if the horizontal distance needed is large. I f so (1/2) m v^2 will decrease and mgh will increase until the ball stops rising at the top and falls, increasing v as it falls. Then by the bottom m g h = 1/2 m v^2, the final kinetic energy due to v is equal to the original m g h. The kinetic energy due to horizontal speed u never changed during the trip except for warming the air a little.
The pitch is thrown with horizontal speed u and vertical speed v.
The kinetic energy due to the horizontal speed (1/2) m u^2 remains constant until ground except for air friction.
However v is tricky because it might be initially upward if the horizontal distance needed is large. I f so (1/2) m v^2 will decrease and mgh will increase until the ball stops rising at the top and falls, increasing v as it falls. Then by the bottom m g h = 1/2 m v^2, the final kinetic energy due to v is equal to the original m g h. The kinetic energy due to horizontal speed u never changed during the trip except for warming the air a little.
Pls help me out of this problem
1 answer