Answer to Question #221773 in Mechanics | Relativity for Hussain Darboe

Question

A firefighter stands 50m from a building that caught fire at a particular height. He starts shooting water to remove the blaze. The water moves at a speed of 40m/s and the pipe is inclined or projected at 30 degrees above the horizontal. Calculate the height where the fire caught?

Answer 1

$v = 40\frac{m}{s}$

$s = 50 m$

$\alpha = 30\degree$

$\vec g = 9.8 \frac{m}{s^2}$

$\text{Consider the movement of a stream of water as}$

$\text{a body thrown at an angle to the horizon:}$

$s(t) = v_xt;$

$v_x= v\cos\alpha$ ;

$t = \frac{s(t)}{ v\cos\alpha}= \frac{50}{ 40\cos30\degree}=1.44s$

$h(t) = v_yt-\frac{gt^2}{2}$

$v_y= v\sin\alpha$

$h(t) = v\sin\alpha *t-\frac{gt^2}{2}=40\sin30\degree=9.83m$

$\text{Answer: } h = 9.83m$