Phy Q.難! (about v and h)

The answer is the 2nd graph, i.e. option B

The equation describing the motion, by conservation of mechanical energy, is:

(1/2)mv^2 = mgh
i.e. v^2 = 2g.h
or v = sqrt[2g].sqrt[h] (where sqrt = square-root)
hence v = k.sqrt(h), where k is a constant
Thus, v increases as h increases.

The slope of the curve is given by,
dv/dh = (k/2).h^(-1/2)
hence, the slope decreases with increasing h

It is clear that option B is the answer.

2012-01-07 12:38:08 補充：
Use the equation of motion: v^2 = u^2 + 2a.s
with u = 0, s = h/sin(theta), (theta) being the angle of slope
thus, v^2 = [2a/sin(theta)].h
or v^2 = K.h, K is a constant
The argument then follows from answer given above.

Thanks for 天同 ! I understand !

我學校有個勁人話B
我本來都覺得係C,但係唔知答案係咩 = ="

2012-01-06 19:40:44 補充：
Can you write(type) your answer?
I want to know!