Question

A fire fighting crew uses a water cannon that shoots water at 25.0m/s at a fixed angle of$\mathbf{53}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal. The firefighters want to direct the water at a blaze that is above ground level. How far from the building should they position their cannon? There are two possibilities; can you get them both?

Short answer

$$\begin{gathered} s_1=8.8m \\ {s}_2=52.4m \end{gathered}$$

Step-by-step solution

Newton’s law

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.

For the vertical and horizontal motion the velocity component will be respectively.

Given

Angle =$53.0°$

Water shoot velocity = 25m/s

Vertical distance of blaze = 10.0m

Position of cannon

From vertical motion,

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 10.0m=\left(20.0m/s\right)t-\frac{1}{2}\times 9.8m/s^2\times t^2 \\ t=0.59sor3.49s \end{gathered}$$

There are two times present, from the horizontal motion equation with using both the time,

Here gravity will be zero,

$$\begin{gathered} s_1=u{t}_1 \\ {s}_1=15m/s\times 0.59s=8.8m \\ {s}_2=u{t}_2 \\ {s}_2=15m/s\times 3.49s=52.4m \end{gathered}$$

When the cannon will be at 8.8m from the building the water hits this spot on the wall and 52.4 m is maximum height passed by it.