Question

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

Answer 1

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 :

`\boxed{\sf{\bold{x_(max) = ((v_0)^2 \cdot \sin(2 \theta))/(g)}}}`

With the following condition :

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

Problem Solving :

We know that :

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

What was asked :

• 
  `\sf{\theta}` = elevation angle = ... °

Step by Step :

• Find the equation value
  `\sf{\bold{theta}}` (elevation angle)

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

`\sf{x_(max) \cdot g = (v_0)^2 \cdot \sin(2 \theta)}`

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

`\sf{(2.5 \cdot 9.8)/((6.8)^2) = \sin(2 \theta)}`

`\sf{(2.5 \cdot 9.8)/((6.8)^2) = \sin(2 \theta)}`

`\sf{(24.5)/(46.24) = \sin(2 \theta)}`

`\sf{\sin(2 \theta) \approx 0.53}`

`\sf{\cancel{\sin}(2 \theta) \approx \cancel{\sin}(32^o)}`

• 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

`\sf{2 \theta = 32^o + k \cdot 360^o}`

`\sf{\theta = 16^o + k \cdot 180^o}`

→
`\sf{k = 0 \rightarrow 16^o + 0 = 16^o}` (T)

→
`\sf{k = 1 \rightarrow 16^o + 180^o = 196^o}` (F)

Second Probability

`\sf{2 \theta = (180^o - 32^o) + k \cdot 360^o}`

`\sf{2 \theta = 148^o + k \cdot 360^o}`

`\sf{\theta = 74^o + k \cdot 180^o}`

→
`\sf{k = 0 \rightarrow 74^o + 0 = 74^o}` (T)

→
`\sf{k = 1 \rightarrow 74^o + 180^o = 254^o}` (F)

`\boxed{\sf{\therefore \theta \{16^o , 74^o\} }}`

Conclusion

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