Answer :
Answer:
8.8 m and 52.5 m
Explanation:
The vertical component and horizontal component of water velocity leaving the hose are
[tex]v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s[/tex]
[tex]v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s[/tex]
Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level
[tex] s = v_vt + gt^2/2[/tex]
[tex]10 = 19.97t - 9.8t^2/2[/tex]
[tex]4.9t^2 - 19.97t + 10 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}[/tex]
[tex]t= \frac{19.9658877511823\pm14.24}{9.8}[/tex]
t = 3.49 or t = 0.58
We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down
t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building
[tex]s_1 = v_ht_1 = 15*0.58 = 8.8 m[/tex]
[tex]s_2 = v_ht_2 = 15*3.49 = 52.5m[/tex]