Question

The velocity v and maximum height h of the water being pumped into the air are related by the equation v=\sqrt(2gh) where g is the acceleration due to gravity (32 feet/second^2). a. Determine the equation that will give the maximum height of the water as a function of its velocity b. The Mayfield Fire Department must purchase a pump that is powerful enough to propel water 80 feet in the air. Will a pump that is advertised to project water with a velocity of 75 feet/second meet the fire department’s needs? Explain.

Answer 1

The velocity v and maximum height h of the water being pumped into the air are related by the equation

v= $\sqrt{2gh}$

where g = 32

(a) To find the equation that will give the maximum height of the water , solve the equation for h

v= $\sqrt{2gh}$

Take square root on both sides

$v^2$ = 2gh

Divide by 2g on both sides

$\frac{v^2}{2g}$ = h

So maximum height of the water h = $\frac{v^2}{2g}$

(b) Maximum height h= 80

velocity v= 75 ft/sec

Given g = 32

h = $\frac{v^2}{2g}$

h = $\frac{75^2}{2*32}$

h= 87.89 ft

The pump withe the velocity of 75 ft/sec reaches the maximum height of 87.89 feet. 87.86 is greater than the maximum height 80 feet.

So the pump will meet the fire department needs.