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A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the water land 2.5 m away

Sagot :

MDKPfromIDN

So, based on the angle values that have been found, the angle of elevation of the nozzle can be 16° or 74°.

Introduction

Hi ! This question can be solved using the principle of parabolic motion. Remember ! When the object is moving parabolic, the object has two points, namely the highest point (where the resultant velocity is 0 m/s in a very short time) and the farthest point (has the resultant velocity equal to the initial velocity). At the farthest distance, the object will move with the following equation :

[tex] \boxed{\sf{\bold{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}}} [/tex]

With the following condition :

  • [tex] \sf{x_{max}} [/tex] = the farthest distance of the parabolic movement (m)
  • [tex] \sf{v_0} [/tex] = initial speed (m/s)
  • [tex] \sf{\theta} [/tex] = elevation angle (°)
  • g = acceleration due to gravity (m/s²)

Problem Solving :

We know that :

  • [tex] \sf{x_{max}} [/tex] = the farthest distance of the parabolic movement = 2.5 m
  • [tex] \sf{v_0} [/tex] = initial speed = 6.8 m/s
  • g = acceleration due to gravity = 9.8 m/s²

What was asked :

  • [tex] \sf{\theta} [/tex] = elevation angle = ... °

Step by Step :

  • Find the equation value [tex] \sf{\bold{theta}} [/tex] (elevation angle)

[tex] \sf{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}} [/tex]

[tex] \sf{x_{max} \cdot g = (v_0)^2 \cdot \sin(2 \theta)} [/tex]

[tex] \sf{\frac{x_{max} \cdot g}{(v_0)^2} = \sin(2 \theta)} [/tex]

[tex] \sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)} [/tex]

[tex] \sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)} [/tex]

[tex] \sf{\frac{24.5}{46.24} = \sin(2 \theta)} [/tex]

[tex] \sf{\sin(2 \theta) \approx 0.53} [/tex]

[tex] \sf{\cancel{\sin}(2 \theta) \approx \cancel{\sin}(32^o)} [/tex]

  • Find the angle value of the equation by using trigonometric equations. Provided that the parabolic motion has an angle of elevation 0° ≤ x ≤ 90°.

First Probability

[tex] \sf{2 \theta = 32^o + k \cdot 360^o} [/tex]

[tex] \sf{\theta = 16^o + k \cdot 180^o} [/tex]

→ [tex] \sf{k = 0 \rightarrow 16^o + 0 = 16^o} [/tex] (T)

→ [tex] \sf{k = 1 \rightarrow 16^o + 180^o = 196^o} [/tex] (F)

Second Probability

[tex] \sf{2 \theta = (180^o - 32^o) + k \cdot 360^o} [/tex]

[tex] \sf{2 \theta = 148^o + k \cdot 360^o} [/tex]

[tex] \sf{\theta = 74^o + k \cdot 180^o} [/tex]

→ [tex] \sf{k = 0 \rightarrow 74^o + 0 = 74^o} [/tex] (T)

→ [tex] \sf{k = 1 \rightarrow 74^o + 180^o = 254^o} [/tex] (F)

[tex] \boxed{\sf{\therefore \theta \{16^o , 74^o\} }}[/tex]

Conclusion

So, based on the angle values that have been found, the angle of elevation of the nozzle can be 16° or 74°.

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