Volcanic Lava Fountains (Derivatives)?

Question

Although the November 1949 Kilauea Iki eruption on the island of Hawaii bean with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater's floor, which at one point shot lava 1900 ft straight into the air (a world record). 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 = (v0)(t) - 16t^2 feet. Begin by finding the time at which ds/dt = 0. Neglect air resistance.)

PS: please use derivatives and show step by step! thanks!

i can't decide whether v0 should be a constant or a variable. i think it should be a constant, so i treated it as such and came up with "s' = v0 - 32t" for the velocity function. but now i can't figure out how to solve for either variable.

Answer 1

It's funny, I'm in the Big Island of Hawaii for the summer and I just hiked along the Kilauea Iki (now dormant) a few days ago.

v0 is supposed to be a constant.
It is the exit velocity you are looking for, and you can assume it stays constant if the height of the fountain doesnot change.
The derivative is ds/dt = v0 - 32 * t, as you found out yourself.

Now let's go back to the physics : the lava reaches the apex of the fountain when its velocity is null. That gives you a relation between v0 and ta, time after which the lava reaches the apex :
ds/dt = 0 = v0 - 32*ta (1)
And you also have data about the ultimate height of the fountain, given by :
ha = v0*ta - 16*ta^2 = 1900 ft (2)

With these two equations, you can solve for v0 (and ta) :
(1) => v0 = 32*ta => ta = v0/32
(2) => v0*ta - 16*ta*ta = 1900
By replacing ta you find :
1/32*v0^2 - 1/64*v0^2 = 1900
v0^2 = 1900 * 64
So v0 = 348 ft per sec is your answer.

Answer 2

In this case, v0 is a constant because the exit velocity can only have been 1 number.

The lava follows a parabolic path

s(t) = v0*t -16*t^2 ft

We know the maximum height was 1900 ft, if we put this in for s(t) at the max time t_max, we cannot solve it because we still don't know v0. However, we know that at the maximum height, the vertical velocity was zero (draw the parabola and see for yourself). so

ds/dt = v0 - 32*t

at t_max this is zero,

0 = v0 - 32*t_max => t_max = v0/32
substitute this into the function for s(t_max)

s(t_max) = 1900 ft = v0*v0/32 - 16*(v0/32)^2

v0^2/32*(1-16/32) = 1900

v0^2 = 1900*2*32 = 121600 (ft/s)^2

... v0 = 348.7 ft/s 5280 ft = 1 mile

v0 = .06 mph

Answer 3

v0 is a constant.
s = dv/dt so s = v0t-16t^2 +c when t = 0 s=0 so c=0
ds/dt = -32t +v0
Set this = 0 and get t = v0/32
So 1900 = v0^2/32 - 16v0^2/32^2
1900*32^2 =32v0^2 -16v0^2
v0 = sqrt(1900*32^2/16) = 80sqrt19 ft/sec