Applications for Derivatives (velocity)

• November 6th 2007, 09:25 PM
Truthbetold
Applications for Derivatives (velocity)
*insert long descriptive paragraph about a volcano*
Paraphrase: With regards to a shot of lava at a point 1900 ft shooting straight into the air, what was the lava's exit velocity in feet per second? in miles per hour?

[HInt: If v0 is the exit velocity of a particle of lava, its height t seconds later will be s= v0t - 16t^2 feet. Begin by finding the time at which ds/dt= 0. Neglect air resistance]

I assume v0 is just another term for the first velocity.

ds/dt= v0 - 32t

v0 - 32t = 0

v0 = 32t

What now? or did I misunderstand?

Thanks!
• November 6th 2007, 09:54 PM
Soroban
Hello, Truthbetold!

Quote:

With regards to a shot of lava at a point 1900 ft shooting straight into the air,
what was the lava's exit velocity in feet per second? in miles per hour?

Hint: If $v_o$ is the exit velocity of a particle of lava,
its height $t$ seconds later will be: . $s \:= \:v_ot - 16t^2$ feet.
Begin by finding the time at which $\frac{ds}{dt} \,= \,0$

I assume $v_o$ is just another term for the first velocity. .
. . . Yes!

$\frac{ds}{dt} \:= \:v_o - 32t \:=\:0\quad\Rightarrow\quad v_o \:=\:32t$
. . .
You were well on your way . . .

You found that the lava reached maximum height at: . $t \,=\,\frac{v_o}{32}$ seconds.

Then the maximum height is: . $s \;=\;v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \;=\;\frac{v_o^2}{64}$ feet

But we are told that the maximum height is 1900 feet.

. . So we have: . $\frac{v_o^2}{64} \:=\:1900$

Got it?