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A space probe is shot upward from the earth. If air resistance is disregarded, a differential equation for the velocity after burnout is v dv/dt = ky^2, where y is the distance from the center of the earth and k is a positive constant. If y0 is the distance from the center of the earth at burnout and v0 is the corresponding velocity, express v as a function of y.
 one year ago
 one year ago
A space probe is shot upward from the earth. If air resistance is disregarded, a differential equation for the velocity after burnout is v dv/dt = ky^2, where y is the distance from the center of the earth and k is a positive constant. If y0 is the distance from the center of the earth at burnout and v0 is the corresponding velocity, express v as a function of y.
 one year ago
 one year ago

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hal_stirrupBest ResponseYou've already chosen the best response.0
write the equation again am not sure what am reading is right.
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
\[v \frac{ dv }{ dt } = ky ^{2} \]
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}Vdv = \int\limits_{}^{} ky ^{2}dy\]\[\frac{ 1 }{ 2 }V ^{2}=ky ^{1} + c\]\[V=\sqrt{2k \frac{ 1 }{ y } + c}\] So having done that, does c turn in to the initial velocity?
 one year ago

phiBest ResponseYou've already chosen the best response.1
except that it is dt not dy I was thinking V= dy/dt so you could say dt= dy/V and sub in for dt in the equation.
 one year ago

phiBest ResponseYou've already chosen the best response.1
that makes it easier. to find c, plug in V0 for V and Y0 for y in your 2nd equation and solve for c
 one year ago

ieatpiBest ResponseYou've already chosen the best response.0
ok I'll give that a shot, thanks!
 one year ago
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