Question

A firefighter, a distancefrom a burning building, directs a stream of water from a fire hose at angleabove the horizontal as shown in Figure. If the initial speed of the stream is, at what heightdoes the water strike the building?

Short answer

$h=d.\tan {\theta }_i-\frac{d^2.g}{2.v_i^2.{\cos }^2{\theta }_i}$

Step-by-step solution

Given and required

Given:

Horizontal distance: d

Initial angle:${\theta }_i$

Initial speed:$v_i$

Required:

Height at which water strikes the building

Projectile motion 

The motion of an item hurled (projected) into the air is known as projectile motion. After the initial force that propels the item into the air, it is only subjected to the pull of gravity.

The movement of water is the movement of projectiles. Therefore, the movement of water can be decomposed into axes and axis movements and observed separately.

Equations:

The x-axis motion of the mug can be expressed as:

$x=v_i.\cos {\theta }_i.t$

Where,

$v_i$is the initial velocity,$t$is the time and${\theta }_i$ is the initial angle.

The y -axis motion of the mug can be expressed as:

$y=y_0+v_i.\sin {\theta }_i.t-\frac{1}{2}.g.t^2$

Where $y_0$is the initial height and $g=9.81\frac{m}{s^2}$.

Step 3: Calculation

The initial height in this problem is $y_0=0$. Therefore, equations$\left(1\right)$and$\left(2\right)$become:

$$\begin{gathered} x=v_i.\cos {\theta }_i.t \\ y=v_i.\sin {\theta }_i.t-\frac{1}{2}.g.t^2 \end{gathered}$$

The maximum distance isrole="math" localid="1663745231112" $x_{max}=d$and the final height is $y=h$:

$$\begin{gathered} d=v_i.\cos {\theta }_i.t_{max} \\ h=v_i.\sin {\theta }_i.t_{max}-\frac{1}{2}.g.t_{max}^2 \end{gathered}$$

Rewriting equations

$$\begin{gathered} t_{max}=\frac{d}{v_i.\cos {\theta }_i} \\ h=v_i.\sin {\theta }_i.t_{max}-\frac{1}{2}g.t_{max}^2 \end{gathered}$$

Step 4: Substituting the values

Substitute the first equation into the second one:

$$\begin{gathered} h=v_i.\sin {\theta }_i.\frac{d}{v_i.\cos {\theta }_i}-\frac{1}{2}.g.\frac{d^2}{v_i^2.{\cos }^2{\theta }_i} \\ h=d.\tan {\theta }_i-\frac{d^2.g}{2.v_i^2.{\cos }^2{\theta }_i} \end{gathered}$$