Question

A water hose is used to fill a large cylindrical storage tank of

Answer 1

Answer:

maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$

Explanation:

Let the nozzle of the hose be at the origin. Then the nearest part of the rim of the tank is at (, ) =  (6, 2) and the furthest part of the rim is at (, ) = (7, 2).

The trajectory of the water can be found as follows:

$x = (v_o cos45) t$

$y = (v_o sin45) t - \frac{1}{2}gt^2$

Now from above two equations we have

$y = x - \frac{gx^2}{v_o^2}$

now we know that height of the cylinder is 2D so we have

$x - \frac{gx^2}{v_o^2} = 2D$

by solving above equation we have

$x = \frac{v_o^2 \pm v_o^2\sqrt{1 - \frac{8gD}{v_o^2}}}{2g}$

now we know that maximum value of x is 7D

so the maximum possible speed by solving above equation for 7D is

$v_{max} = \sqrt{\frac{49}{5}gD}$

minimum possible value of speed for solving x = 6D is given as

$v_{min} = \sqrt{9gD}$