projectile

a)
For the ball travel from C to the net, time taken is t1
Vertical displacment = 0.4 - 0.15 = 1/2 * g * t1^2
0.25 = 1/2 * 10 * t1^2
t1 = 0.2236s
Horizontal displacement = 1.35 = v * t1
v = 6.037 m/s

b)
For the ball travel from C to D, time taken is t2
Vertical displacment = 0.4 = 1/2 * g * t2^2
0.4 = 1/2 * 10 * t2^2
t2 = 0.2828s
Horizontal displacement = v * t2 = 6.037 * 0.2828 = 1.707 m
Horizontal speed = 6.037 m/s
Vertical speed = g * t2 = 10 * 0.2828 =2.828 m/s
Speed of the ball just before it hits the table
= (speed vert^2 + speed horiz^2)^(1/2)
= (6.037^2 + 2.828^2)^(1/2)
= 6.666 m/s

the ans of a may be wrong.
a. t for down (.4-.15) =0.25m = 1/2(10)t^2 , therefore t =0.5s
Horizontally, S=vt=1.35=v(0.5)
therefore v=2.7m/s
b. when s = 0.4(just before hit D), vertical velocity = s = 0.4=1/2(10)t^2, therefore t^=0.08
velocity just b4 hit = square root of Vv^2 + Vh^2, Vh=2.7, Vv=10t.
therefore 2.7^2 +100t^2 = 7.29 + 8 =15.29. i.e. Vb=3.91m/s

2007-08-24 15:01:23 補充：
Very Sorry, I made a big mistake!