Question

A firefighting crew uses a water cannon that shoots water at 30.0 m/s at a fixed angle of 47.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 12.0 m above ground level.

How far from the building should they position their cannon? There are two possibilities (d1 < d2 ); can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

Answer 1

Step-by-step explanation:

Projectile Motion

When an object is launched with an initial velocity and angle above the horizontal, it describes a curve called a parabola, with a vertex in the point where the object has its maximum height and then it finally returns to the ground level.

The water is to be shot to a fixed height from a horizontal distance to be determined. Please refer to the figure below. There are two possible solutions to the problem: the first when the water is still moving upwards, and the second when it has already reached the maximum height and is going downwards. The distances are called d1 and d2.

The height of an object launched at vo and \theta is

`\displaystyle y=V_o\ sin\theta \ t-(g\ t^2)/(2)`

And the horizontal distance it reaches can be computed by

`\displaystyle d=V_o\ cos\theta \ t`

Let's solve it for t as follows

`\displaystyle t=(d)/(V_o\ cos\theta)`

And replace it on the previous equation

`\displaystyle y=V_o\ sin\theta \left((d)/(V_o\ cos\theta)\right)-(g)/(2)\left((d)/(V_o\ cos\theta)\right)^2`

Simplifying and rearranging, we have the equation that relates both components of the combined motion

`\displaystyle y=d\ tan\theta -(gd^2)/(2\ V_o^2\ cos^2\theta)`

Plugging in the given values

`\displaystyle y=12\ ,\ V_o=30,\ \theta =47^o`

We have

`\displaystyle 12=d\ tan47^o-(9.8\ d^2)/(2*30^2\ cos^247^o)`

Rearranging

`\displaystyle 12=1.0723\ d\ -0.0117\ d^2`

We have to solve the second-degree equation

`\displaystyle 0.0117\ d^2-1.023\ d+12=0`

There are two feasible solutions:

`\displaystyle d_1=13.05\ m`

`\displaystyle d_2=78.56\ m`