When fighting a fire the velocity V of water being pumped into the air is modeled by the function v=�ã2hg, where h represents the maximum height of the water and g represents the acceleration due to gravity (32 ft/s^2)
Solve the function for h.
Would this be v=�ã(2h*32)
The Jackson Fire dept must purchase a pump that will propel water 90ft into the air. will a pump advertised to project water with a velocity of 77 ft/sec meet the dept need? explain
The square root symbol did not show using alt+251.
v=�ã2hg
The alt+251 would probably work for Microsoft Word or other word processors. Here the symbols require unicode sequences.
See for example:
http://tlt.its.psu.edu/suggestions/international/bylanguage/mathchart.html
You can show the square-root symbol using the following letter sequence (skip the spaces in-between):
& r a d i c ;
to get the symbol √.
Thank you MathMate for the instructions on the √
"Solve the function for h. "
requires that h will be the dependent variable (similar to y) in terms of v.
Given:
v=√(2gh)
square both sides:
v² = 2gh
solve for h as a function of g:
h(v)=v²/(2g)
For the new purchase,
h(77)=77²/(2*32.2)
= 92.1 ft.
> minimum requirement of 90 ft.
Note: the calculated maximum h will be attained if the water is aimed at 45° with the horizontal, and without air-resistance.
To solve the function v=√(2hg) for h, we need to isolate h.
Given that v=√(2hg), we can square both sides of the equation to remove the square root:
v^2 = 2hg
Now, divide both sides by 2g:
v^2 / (2g) = h
So, the equation solved for h is h = v^2 / (2g).
Now let's apply this equation to the second question.
The pump is advertised to project water with a velocity of 77 ft/sec. We can use the given value of v=77 ft/sec and substitute it into our equation:
h = (77 ft/sec)^2 / (2 * 32 ft/s^2)
Calculating this, we get:
h = 592.15625 ft^2/s^2 / 64 ft/s^2
Simplifying further:
h ≈ 9.253 ft
Therefore, according to these calculations, the pump advertised to project water with a velocity of 77 ft/sec would meet the Jackson Fire Department's need, as it can propel water approximately to a maximum height of 9.253 ft, which is less than their requirement of 90 ft.