Posted by Anonymous on Friday, March 14, 2008 at 3:16pm.
I answered this about two days ago. Please check previous postings. If you need additional help, show your work.
thanks! Is the answer supposed to be -30m/s for Vperp, though? Since it's going down into the tube?
Also I don't know how to find time without distance.
The only way to do it can be
(V2-V1)/a
30-40/-10
(*we are supposed to use 10 for acceleration)
= 1s
I'm not sure if that's right.
d)
Would you multiply the perpendicular component of V2 by the time found for the height?
there are of course two possible solutions. It could be at 45 degrees on the way up or on the way down.
Your comment that it goes DOWN into the tube means that indeed your vertical component is - 30 while your horizontal component remains +30
Now
v = vertical speed = Vo - g t
so
-30 = 40 - 10 t
10 t = 70
t = 7 seconds to tube entry
Now where did it go in those seven seconds
x = 30 t = 30 * 7 = 210 meters
h = height = Vo t + (1/2) a t^2
but a = -10 m/s^2
h = 40*7 -5*49
h = 280-245
h = 35 meters
hi
1tdc03sz4jr9fith
good luck
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