Find velocity required to reach a certain height given only

[haiku11]

On the surface of the sun the acceleration due to gravity is approx 0.25km/s². A mass of gas forming a solar prominence rises from the sun's surface. If only gravity is considered, what must its initial upward velocity be, if it is to reach a height of 24000km above the surface?

I can't figure out how to do this, I have:
a(t) = 0.25 the integral would give me:
v(t) = 0.25t + C the integral of this would give me:
s(t) = 0.125t² + Ct + D

I'm assuming D would be 0 because that's where the prominence starts on the ground so the equation becomes:
24000 = 0.125t² + Ct
0 = 0.125t² + Ct - 24000

I don't know where to go from here because there are 2 variables and I can't do any substitution using the previous equations. Trying to use the quadratic formula made everything really messy when trying to rearrange it in terms of "t". I even tried doing this the physics way although this is a calculus problem and I still couldn't do it with the 5th motion equation because I would be trying to take the square root of a negative. But if I ignore the negative and take the root I get the approximately right answer of around 109km/s.

 

[tkhunny]

Re: Find velocity required to reach a certain height given o

Two items:

1) INITIAL velocity is associated with t = 0.
2) Acceleration due to gravity probably should be negative.

 

[soroban]

Re: Find velocity required to reach a certain height given o

Hello, haiku11!

On the surface of the sun the acceleration due to gravity is approx 0.25 km/s².
A mass of gas forming a solar prominence rises from the sun's surface.
If only gravity is considered, what must its initial upward velocity be
if it is to reach a height of 24,000 km above the surface?

Your preliminary work is quite good . . .

\(\displaystyle \text{The acceleration is: }\:a(t) \:=\:-\tfrac{1}{4}\)

\(\displaystyle \text{The velocity is: }\:v(t) \:=\:-\tfrac{1}{4}t + v_o\)

\(\displaystyle \text{The height is: }\:s(t) \:=\:-\tfrac{1}{8}t^2 + v_ot + h_o\)

\(\displaystyle \text{Since the prominence starts at "ground level", }h_o = 0.\)
. . \(\displaystyle \text{Hence, the height function is: }\:s(t) \:=\:v_ot - \tfrac{1}{8}t^2\)


\(\displaystyle \text{Assuming that 24,000 km is the }maximum\text{ height of the prominence,}\)
. . \(\displaystyle \text{this occurs when }v(t) = 0.\)

\(\displaystyle \text{We have: }\:-\tfrac{1}{4}t + v_o \:=\:0 \quad\Rightarrow\quad t \:=\:4v_o\)
. . \(\displaystyle \text{Then: }\:s(\text{max}) \:=\:v_o(4v_o) -\tfrac{1}{8}(4v_o)^2 \:=\:2v_o^2\)


\(\displaystyle \text{Since }s(\text{max}) \,=\,24,\!000\)
. . \(\displaystyle \text{we have: }\:2v_o^2 \:=\:24,\!000 \qud\Rightarrow\quad v_o^2 \:=\:12,\!000 \quad\Rightarrow\quad v_o \:=\:\sqrt{12,\!000}\)

. . \(\displaystyle \text{Therefore: }\:v_o \;=\;20\sqrt{30} \;\approx\;109.5\text{ km/sec.}\)