Question

The water from a fire hose is directed at a building . The water leaves the pipe at a speed of 25 m/s at an angle of 60 deg to the horizontal. If the building is 45 m away, at what height does it hit the building ?

Answer 1

Approximately
`14\; \rm m` above the height of the fire hose, assuming that air resistance is negligible and that
`g = -9.81\; \rm m \cdot s^(-2)`.

Step-by-step explanation:

Consider the motion of water particles in two directions: vertical and horizontal.

Assuming that air resistance on the water particles is negligible.

• Vertically, water particles would accelerate at
  `(-9.81\; \rm m \cdot s^(-2))` (towards the ground.)
• Horizontally, water particles would travel at a constant velocity.

Initial velocity:
`v_0 = 25\; \rm m \cdot s^(-1)`. Angle of elevation:
`\epsilon = 60^\circ`. Calculate the initial velocity of these water particles in these two components:

• Initial horizontal velocity:
  `v_0 (\text{horizontal}) = v_0 \cdot \cos\left(\epsilon\right) \approx 12.5\; \rm m \cdot s^(-1)`.
• Initial vertical velocity:
  `v_0 (\text{vertical}) = v_0 \cdot \sin\left(\epsilon\right) \approx 21.7\; \rm m \cdot s^(-1)`.

How much time would it take for a water particle reaches the building from the hose? That particle needs to travel
`x(\text{horizontal}) = 45\; \rm m` at a constant horizontal speed of
`v(\text{horizontal}) \approx 12.5\; \rm m \cdot s^(-1)`. Therefore:

`\displaystyle t = \frac{x(\text{horizontal})}{v(\text{horizontal})} \approx 3.6\; \rm s`.

In other words, that water particle would be approximately
`3.6\; \rm s` into its flight when it hits the building. What would be its height? Assuming that the air resistance on that particle is negligible. The height of that water particle at time
`t` may be modeled using the SUVAT equation:

`\displaystyle x(\text{vertical}) = (1)/(2)\, a \, t^(2) + \left[v_0(\text{vertical})\right]\, t`,

where:

• 
  `x(\text{vertical})` gives the height of the water particle (relative to where it was launched.)

• 
  `a = g = -9.81\; \rm m \cdot s^(-2)` (negative because gravitational acceleration points towards the ground.)

• 
  `v_0(\text{vertical}) \approx 21.7\; \rm m \cdot s^(-1)` (see above.)

• 
  `t \approx 3.6\; \rm s` (that's the time when the water particle hits the building.)

Calculate the height of the water particle when it hits the building:

`\begin{aligned}x(\text{vertical}) &= (1)/(2)\, a \, t^(2) + \left[v_0(\text{vertical})\right]\, t \\ &\approx (1)/(2) * \left(-9.81\; \rm m \cdot s^(-2)\right) * (3.6\; \rm s)^2 + 21.7\; \rm m \cdot s^(-1) * 3.6\; \rm s \\ &\approx 14\; \rm m\end{aligned}`.