Question

A fire fighters crew uses water cannon at 25m/s at a fixed angle of 53.0 degree above horizontal. The fire fighters want to direct the water to a blaze that is 10.0m above ground level. How far from the blaze should they position their cannon?

Answer 1

Explanations & Calculations

• Careful inspection of the situation along with the motion equation $\small (x-x_0 )=v_0t+\frac{1}{2}at^2$ express that there exist two sets of values for the time that correspond to a given displacement.
• Then, applying this to the given situation yields,

$\qquad\qquad \begin{aligned} \small \uparrow10m&=\small 25\sin53 t+\frac{1}{2}\times(-9.8ms^{-2})t^2\\ \small t&=\begin{cases} \small 3.489\,s\\ \small 0.585\,s \end{cases} \end{aligned}$

• It should be the larger value that fits the given situation as the trajectory of the water column is a projectile.

• Then we can calculate from how far the water jet should be aimed.

$\qquad\qquad \begin{aligned} \small \to (x-x_0)&=\small v_0 t\\ \small x&=\small 25\cos 53\times 3.489\\ &=\small \bold{52.493\,m} \end{aligned}$