Question

Old Faithful Watching Old Faithful erupt, you notice that it takes a time t for water to emerge from the base of the geyser and reach its maximum height. (a) What is the height of the geyser, and (b) what is the initial speed of the water? Evaluate your expressions for (c) the height and (d) the initial speed for a measured time of 1.65 s.

Answer 1

Part A. Using 2nd kinematic equation in vertical direction:

y = U*t + (1/2)*a*t^2

U = Initial speed = V₀

t = time taken to reach max height

a = acceleration due to gravity = -g (-ve sign for downward direction)

y = max height reached = h

So,

h = V₀*t + (1/2)*(-g)*t^2

h = V₀*t - (1/2)*g*t^2

Part B.

Using 1st kinematic equation in vertical direction:

V = U + a*t

V = final velocity at max height = 0, So

0 = V₀ + (-g)*t

V₀ = g*t

Part C

When t = 1.65 sec

g = 9.81 m/s^2

So,

V₀ = g*t = 9.81*1.65 = 16.1865

V₀ = 16.2 m/sec

Now max height will be:

h = V₀*t - (1/2)*g*t^2

h = 16.1865*1.65 - (1/2)*9.81*1.65^2 = 13.3538 m

h = max height = 13.4 m

Part D.

When t = 1.65 sec

g = 9.81 m/s^2

So,

V₀ = g*t = 9.81*1.65 = 16.1865

V₀ = 16.2 m/sec = Initial speed

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