Water leaves a fireman’s hose (held near the ground) with an initial velocity v0=19.5v0=19.5 m/s at an angle θθ= 34 degrees above the horizo

Question

Water leaves a fireman’s hose (held near the ground) with an initial velocity v0=19.5v0=19.5 m/s at an angle θθ= 34 degrees above the horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.
Using v0,θ,v0,θ, and g, write an expression for the time, t max, the water travels to reach its maximum vertical height. At what horizontal distance d from the building base, should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of v0,θ,v0,θ,and g.

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Diễm Kiều 1 year 2021-09-01T08:54:34+00:00 1 Answers 0 views 0

Answers ( )

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    2021-09-01T08:56:08+00:00

    Answer:

    tmax = 1.11 s, d = 35.98 m

    Explanation:

    Here is the complete question

    Water leaves a fireman’s hose (held near the ground) with an initial velocity  

    v
    0
    =
    19.5  m/s at an angle  θ
    = 34 degrees above the horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.

    Using  v
    0
    ,
    θ
    ,
    and g, write an expression for the time, t max, the water travels to reach its maximum vertical height. At what horizontal distance d from the building base, should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of  v
    0
    ,
    θ
    ,
    and g.

    Solution

    Since the water leaving the hose is considered to be a projectile motion with initial velocity,v₀ = 19.5 m/s and an angle  θ = 34. The time tmax it takes the water to reach maximum height is given by

    tmax = v₀sin
    θ/g = 19.5 × sin34/9.8 = 19.5 × 0.5592/9.8 = 10.904/9.8 = 1.113 s ≅ 1.11 s

    The horizontal distance,d from the base of the building at which the hose must be placed to reach maximum height is the range of the projectile and is given by

    d = v₀²sin2θ/g = 19.5²sin(2×34)/9.8 = 19.5²sin68/9.8 = 380.25 × 0.9272/9.8 = 352.562/9.8 = 35.98 m

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