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A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 52.0 ∘ 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

User Nico AD
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1 Answer

2 votes

Answer:

49.4 m

Step-by-step explanation:

u = 25 m/s

θ = 52°

y = 12 m

Let the distance between the cannon and the building is d.

Use the general equation of projectile path


y = d tan\theta -(gd^(2))/(2u^(2)Cos^(2)\theta )

By substituting the values, we get


12= d tan52 -\frac{9.8* d^(2)}{2* 25 * 25Cos^(2)52


12=1.28 d -0.021 d^(2)


0.021 d^(2)-1.28d+12= 0


d=(1.28\pm √(1.64-1.008))/(0.042)

d = 49.4 m

User John Kariuki
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