Question

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

Answer 1

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=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