Answer:
maximum possible speed by solving above equation for 7D is
[tex]v_{max} = \sqrt{\frac{49}{5}gD}[/tex]
minimum possible value of speed for solving x = 6D is given as
[tex]v_{min} = \sqrt{9gD}[/tex]
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:
[tex]x = (v_o cos45) t[/tex]
[tex]y = (v_o sin45) t - \frac{1}{2}gt^2[/tex]
Now from above two equations we have
[tex]y = x - \frac{gx^2}{v_o^2}[/tex]
now we know that height of the cylinder is 2D so we have
[tex]x - \frac{gx^2}{v_o^2} = 2D[/tex]
by solving above equation we have
[tex]x = \frac{v_o^2 \pm v_o^2\sqrt{1 - \frac{8gD}{v_o^2}}}{2g}[/tex]
now we know that maximum value of x is 7D
so the maximum possible speed by solving above equation for 7D is
[tex]v_{max} = \sqrt{\frac{49}{5}gD}[/tex]
minimum possible value of speed for solving x = 6D is given as
[tex]v_{min} = \sqrt{9gD}[/tex]
