Question

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 to reach a building 41.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.
Required:
a. Find the angle of elevation of the hose.
b. Find the speed in m/s of the water at the highest point in its trajectory.
c. Find the acceleration in m/s^2 of the water at the highest point in its trajectory.
d. How high above the ground in m does the water strike the building?
e. How fast is it moving in m/s just before it hits the building?

Answer 1

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

• Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
• Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
• vₓ₀ = v * cos θ (1)
• where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
• Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        $x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

• Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        $cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

• ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

• At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
• Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
• vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

• At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

• Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       $\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

• Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
• v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
• Replacing (7) in (6), we get:

       $\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

e)

• When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
• The horizontal component, since it keeps constant, is just v₀x:
• v₀ₓ = 13.7 m/s
• The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       $v_{fy} = v_{oy} - g*t (9)$

• Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       $v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

• Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       $v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$