A geyser sends a blast of boiling water high into the air. During the eruption, the height (h) (in feet) of the water (t)seconds after being forced out from the ground could be modeled by h = -16t*t + 70t.
My question is how do you find the initial velocity of the boiling water?
To find the initial velocity of the boiling water, we need to recall that the initial velocity (often denoted as "u") is the velocity of an object at the beginning of its motion, in other words, at t = 0.
In this scenario, we are given a height function for the water, h = -16t^2 + 70t. To find the initial velocity, we can use the fact that velocity (v) is the rate of change of position with respect to time, or more formally, v = dh/dt (the derivative of the height function with respect to time).
So, to find the initial velocity, we need to differentiate the height function h with respect to time t:
h = -16t^2 + 70t
dh/dt = d(-16t^2)/dt + d(70t)/dt
First, let's differentiate -16t^2 with respect to t:
d(-16t^2)/dt = -32t
Next, let's differentiate 70t with respect to t:
d(70t)/dt = 70
Now, we add the two results together:
dh/dt = -32t + 70
This expression represents the velocity of the boiling water at any given time t. However, we want to find the initial velocity at t = 0.
When t = 0, the term "-32t" becomes 0, and we are left with:
dh/dt = 70
Therefore, the initial velocity is 70 ft/s.
You either need to know calculus or a little physics.
If you have had any calculus you take the derivative , rate of change of height with time, at t = 0
dh/dt = -32 t^2 + 70
at t = 0, that is 70 ft/second
If you have taken physics you know:
if acceleration = a
then velocity = Vo + a t
and distance = intitial position + Vo t + (1/2) a t^2
in this case
a = - 32 ft/s^2 the acceleration of gravity
thus v = Vo - 32 t
and
h = 0 + Vo t - 16 t^2
which is given as h = 70 t - 16 t^2
so Vo = 70