How i can solve this question?
Posted in xn--kfs74mzzid01b.com edit by webmaster on January 7th, 2009
a- What should be m in order to have these two objects to collide ?
b-what is the height of the object when Vo=sqrt(g.L)
Y = 2Vo(sin m)(T) - (1/2)gT^2
where
g = acceleration due to gravity
For the other object vertically launched,,
Y = Vo(T) - (1/2)gT^2
and for the two objects to collide, the 2 equations must be equal, hence
2Vo(sin m)(T) - (1/2)gT^2 = Vo(T) - (1/2)gT^2
and also, the "T's" for both objects are the same.
Simplifying the above,
2Vo(sin m)(T) = Vo(T)
and since the factors "Vo" and "T" appear on both sides of the equation, the above becomes
2(sin m) = 1
sin m = 1/2
m = arc sin 1/2
m = 30 degrees
<< what is the height of the object when Vo=sqrt(g.L) >>
I do not understand what "height" you are referring to here.
They need to have the same height at the same time.
y = yo + vo t + 1/2 a t^2
h1 = 0 + (2vo)(sin m)t + 1/2 (-9.8)t^2
h1 = 2vo sin m t - 4.9t^2
h2 = 0 + vo t + 1/2 (-9.8)t^2
h2 = vo t - 4.9t^2
h1 = h2
2vo sin m t - 4.9t^2 = vo t - 4.9t^2
2vo sin m t = vo t
2 sin m = 1
sin m = 1/2
m = 30 degrees
b)
v^2 - vo^2 = 2ax
0 - vo^2 = -2gh
vo^2/2g = h
gL/2g = h
L/2 = h
b. 45 degrees? Not sure...
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