2006-09-21 19:30:18 · 3 answers · asked by CaliM3 1 in Science & Mathematics ➔ Physics
Let v1=initial velocity, v2=velocity of the object at just starting the 28.5 m section, v3=bottom striking velocity, h1=the height above the 28.5 m section, h2=28.5 m, t1=time to traverse h1, t2= 1.85 s=time to traverse h2. Then,
h2=v2*t2+0.5*g*t2^2
28.5=v2*1.85+0.5*9.81*1.85^2
Thus, v2=6.33 m/s
h1=v2^/(2*g)
h1=6.33^2/(2*9.81)
h1=2.04 meters
v2=g*t1
Thus, t1=v2/g
t1=6.33/9.81
t1=0.65 second
v3=sqrt(2*g*(h1+h2))
v3=sqrt(2*9.81*(2.04+28.5))
v3=24.5 m/s
or, v3=v2+g*t2
v3=6.33+9.81*1.85
v3=24.5 m/s
2006-09-21 20:34:27 · answer #1 · answered by mekaban 3 · 0⤊ 0⤋
What do you want to know? i.e., from what height did the object fall?
2006-09-22 02:33:20 · answer #2 · answered by Double Century Dude 3 · 0⤊ 0⤋
Kewl.
I can see that happening, yes.
Doug
2006-09-22 03:07:28 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋