Question

A firefighting crew uses a water cannon that shoots water at 25m/s at a fixed angle of 53 degrees above the horizontal. The firefighters want to direct the water at a blaze that is 10.0m above ground level. How far from the building should they position their cannon? There are two possibilities; can you get them both?

Answer 1

Projectile motion relations:

$y-y_0=v_{0y}t+ \frac{1}{2}a_yt^2 \;(1)\\ x-x_0=x_{0x}t+ \frac{1}{2}a_xt^2 \;(2)$

where x,y are positions on x/y axis, $x_0, \;y_0$ are initial positions, $v_{0x}, \;v_{0y}$ are initial velocities and $a_x, \;a_y$ are accelerations. We can choose that the positive direction of x-axis is to the right and the positive y-direction is upwards. Problem asks us to find $x-x_0≡ D$ distance that we can position our water cannon if we know height H of the spot that we want to hit $(y-y_0≡H)$ , the initial speed of the cannon, and angle of the water. So, first, we can use equation (1) to find time t

$H=v_0sinα_0t - \frac{1}{2}gt^2 \\ 0=-4.9t^2 + 25 \times sin 53° \times t -10 \\ t_1=3.49 \;s \\ t_2=0.58 \;s$

If we put results into equation (2), we can get distances that we can place our water gun:

$D_1=v_0cosα_0t_1 \\ =25 \times cos53° \times 3.49 = 52.5 \;m \\ D_2 = v_0cosα_0t_2 \\ =25 \times cos53° \times 0.58 = 8.7 \;m$

Water cannon can be placed at distances 52.5 m and 8.7 m.