Q56P - University Physics with Modern Physics

Question

A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D. The hose shoots the water at $\mathbf{45}\mathbf{°}$above the horizontal from the same level as the base of the tank and is a distance 6D away (Fig. P3.56). For what range of launch speeds $\left(v_0\right)$ will the water enter the tank? Ignore air resistance, and express your answer in terms of D and g.

Step-by-Step Solution

Answer

The range of speed will be $3\sqrt{\mathrm{gD}} \mathrm{and} 3.13\sqrt{\mathrm{gD}}$.          

Step 1: Introduction

The velocity contains two components, one is horizontal component and another one is vertical. 

 

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$ 

 

Where, s, u, t, a, are displacement, initial velocity, time and acceleration.

Step 2: Given

Angle = $45°$

Step 3: Explanation

For the vertical motion, gravity is negative

 

$s_v=u_v{t}_v-\frac{1}{2}g{t_v}^2$

 

For the horizontal motion, gravity is zero

             

$s_H=u_H{t}_H$

 

When the water goes in the with the minimum velocity,

                                                                                           

$s_v=2D,s_H=6D$

 

When the water goes in the with the maximum velocity,

 

$s_v=2D,s_H=7D$

 

In both the cases,

 

$\sin 45°=\cos 45°=1/\sqrt{2}$

 

To reach minimum distance,

$$\begin{gathered} 2\mathrm{D}=\frac{\mathrm{u}}{\sqrt{2}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \mathrm{and} 6\mathrm{D}=\frac{\mathrm{u}}{\sqrt{2}}\mathrm{t} \\ 2\mathrm{D}=\frac{\mathrm{u}}{\sqrt{2}}\frac{6\sqrt{2}\mathrm{D}}{\mathrm{u}}\frac{1}{2}\mathrm{g}{\left(\frac{6\sqrt{2}\mathrm{D}}{\mathrm{u}}\right)}^2 \\ \mathrm{u}=3\sqrt{\mathrm{g}}\mathrm{D} \end{gathered}$$ 

 

To reach maximum distance,

$$\begin{gathered} 2D=\frac{u}{\sqrt{2}}t-\frac{1}{2}g{t}^2 and 7D=\frac{u}{\sqrt{2}}t \\ 2D=\frac{u}{\sqrt{2}}\frac{7\sqrt{2}D}{u}-\frac{1}{2}g\left(\frac{7\sqrt{2}D}{u}\right) \\ u=3.13\sqrt{g}D \end{gathered}$$