556556

你沿水平向上30度 擲出一個球，當球達到其最過過度一半時，球多少百分比的初始動能轉為引力勢能？
ans 18.75%
但題解 寫到 u sin30 垂直分量 球達到最高點的一半時 u= u sin30 /2 ,,, 但我計極都諗唔明 個球到一半高的位置的 v 係usin30 /2 ........

我覺得錯，你地點睇？

yeungwk91

when 個球到一半高
垂直分量 = u sin30 / root2
by v_y^2 = u_y^2 + 2as

0 = (u sin30 )^2 + 2 (-g) H
v^2 = (u sin30 )^2 - g H = g H

so
ans = KE ratio ( sin30 / root2)^2 + (cos 30)^2 = 87.5%
so 12.5% KE changes to PE

556556

引用:

原帖由 yeungwk91 於 2016-2-17 05:37 PM 發表
when 個球到一半高
垂直分量 = u sin30 / root2
by v_y^2 = u_y^2 + 2as

0 = (u sin30 )^2 + 2 (-g) H
v^2 = (u sin30 )^2 - g H = g H

so
ans = KE ratio ( sin30 / root2)^2 + (cos 30)^2 = 87.5%
so ...

agree~ 計到都係12.5% , 姐係分mock ans 錯

yeungwk91

which mock is it?

antonyluk

Just to share a quick way to tackle this kind of problem:
Let the total initial kinetic energy be E.
Initial horizontal KE: 3/4 E
Initial vertical KE: 1/4 E

Half of the initial vertical KE is converted into GPE when the height is half of the max.
So GPE = 1/2*1/4 E = 1/8 E = 12.5% of initial KE

[ 本帖最後由 antonyluk 於 2016-2-18 05:49 PM 編輯 ]