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 :

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

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)

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

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

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

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{24.5}{46.24} = \sin(2 \theta)}$

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

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

• 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°.