A fountain rises to 560 ft. what's initial speed of water? how long will it take the water to reach the top?

Question

A fountain rises to 560 ft. what's initial speed of water? how long will it take the water to reach the top?

Answer 1

Firstly, this is one h**l of a fountain!! I'm not even sure that there is a fountain this amazing in the world.

This is quite a simple motion question in 1 dimension.

This is what we know:

displacement = 560 ft = 170.69 m ( I'll be working in metres)

Acceleration = -9.8 ms-1 (the acceleration of gravity)

final velocity = 0 ms-1 (this is when the water is at the top of the stream, it is not moving)

So we can find the initial velocity first using this formula:

v^2 = u^2 + (2*a*s)

(v is final velocity = 0, u is initial velocity, what we are finding, a is acceleration =-9.8ms-1, s is displacement = 170.69 m)

v^2 = u^2 + (2*a*s)

v^2 - (2*a*s) = u^2

u = sq. rt.[v^2 - (2*a*s)]

   = sq. rt.[0^2 - (2x-9.8x170.69)]

   = 57.84 ms-1 (The initial velocity of the water)

   = 190 ft./s

To find time we use this formula:

v = u + a*t

t = (v - u) / a

  = (0 - 57.84) / -9.8

  = 5.9 seconds (this is how long it takes the water to reach the top of the fountain).

Have a good day.