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A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the water land 2.5m away?

Respuesta :

wizard123
The speed of water can be split into vertical and horizontal speed components:
[tex]v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta[/tex]

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
[tex]y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t[/tex]
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
[tex]-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3[/tex]
onyebuchinnaji

The angle the nozzle should be pointed in order to project the water at the given distance is 17.7⁰

The given parameters;

  • velocity of the hose, v = 6.5 m/s
  • range of the projected water, R = 2.5 m

let the angle of projection, = θ

The angle of projection is calculated as follows;

[tex]R = \frac{u^2 sin(2\theta )}{g}[/tex]

where;

g is the acceleration due to gravity = 9.8 m/s²

Substitute the given values and solve for the angle of projection;

[tex]R = \frac{u^2 sin(2\theta )}{g} \\\\sin(2\theta ) = \frac{Rg}{u^2} \\\\sin(2\theta ) = \frac{2.5 \times 9.8}{6.5^2} \\\\sin(2\theta ) = 0.5798\\\\2\theta = sin^{-1} (0.5798) \\\\2\theta = 35.4^0\\\\\theta = \frac{35.4^0}{2} \\\\\theta = 17.7^0[/tex]

Thus, the angle of projection is 17.7⁰

Learn more here: https://brainly.com/question/15502195

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