A company manufactures highpowered fountains that can be use

A company manufactures high-powered fountains that can be used to create elaborate water shows set to light and music. The company makes three models of fountains, each of which can launch a vertical stream of water to a different height, as indicated in the table below.

Time to fall(s)

Complete the following tasks, filling in the missing values in the table as you find them:

Find the time water would take to fall from each height in the table back to its starting level. Round your answer to the nearest thousandth of a second.

Find the initial upward velocity required for fountain B to launch the water to its maximum height. Round your answer to the nearest tenth of a foot per second.

Create a function that will give you the minimum initial upward velocity, v, required to launch water to any given height,h.

Solution

we can use 3 useful equations of motion

1) v = u +at

2) s = ut +(1/2) at²

3)v² - u² =2as

where v = final velocity, u = initial velocity, a =acceleration of particle, s= distance covered , t = time taken

So for our queation when water jet is going upward its final velocity will be zero

hence for ModelA

s =h(t) = 77ft , t = 2.194, u = 70.2, v = 0

using 3rd equation of motion

0 - (70.2)² = 2a *(77)

=> 2a = -(70.2)²/77   = - 64.0005194805        

=> a= -32.0002597403          (it is acceleration due to gravity and will be negative when jet is going upward, +ve when coming donq)

Now for model B) s= h(t) = 100ft

v = 0 in upward motion

3rd eaution of motion will gives

0² - u² = (2a)* 100

=>     u² = -(2a)* 100 = (64.0005194805) *100

=> u = 10 * sqrt(64.0005194805)

         = 10* 8.00003246747 ft/sec

          = 80.0003246747   ft/sec

now to get the time to come back to original position, we have s= h(t) = 100ft, u = 0 (initial velocity will be zero in downward motion)

use 2nd eqution of motion

s = ut +(1/2) at²

100 = 0 * t + (1/2) *32.0002597403 t²                       (a is positive in downward motion)

=> 200/32.0002597403 = t²

=> t =2.49998985396 sec;

and velocity after t sec can be fount using 1st equation

v = u+at

v = 0 + 32.0002597403*2.49998985396

v = 80.0003246747

for modelC, use 2nd eation with u = 0 and s =h(t) = 240ft

240 = 0 + (1/2) *32.0002597403 t²

t² = 280/32.0002597403 = 8.74992897784

=> t= 2.95802788659 sec;

function can be found using 3rd eation of motion with final velocity =0 and initial velocity = v

0 - v² = (-2a)h                           (acceleration due to gravity will be -ve in upward motion)

v = sqrt(2ah)                      is the answer