Water leaves a fireman’s hose (held near the ground) with an initial velocity v0= 22.5 m/s at an angle θ = 28.5° above 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.
a) Using v0, θ, and g, write an expression for the time, tmax, the water travels to reach its maximum vertical height.
b) At what horizontal distance d from the building base, where 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, and g.

To solve this problem, we need to use the principles of projectile motion. We have an initial velocity \( v_0 = 22.5 \, \text{m/s} \) and an angle of projection \( \theta = 28.5^\circ \). The water acts as a projectile moving under the influence of gravity \( g = 9.8 \, \text{m/s}^2 \). Let's go through each part step-by-step.

### a) Time to Reach Maximum Vertical Height \( t_{\text{max}} \)

The time to reach the maximum height for a projectile is when the vertical component of the velocity becomes zero. The vertical component of the initial velocity \( v_{0y} \) is given by:

\[
v_{0y} = v_0 \sin \theta
\]

The vertical velocity at any time \( t \) is given by:

\[
v_y(t) = v_{0y} - g t
\]

At the maximum height, the vertical velocity \( v_y(t) = 0 \). Therefore, we set up the equation:

\[
0 = v_0 \sin \theta - g t_{\text{max}}
\]

Solving for \( t_{\text{max}} \):

\[
t_{\text{max}} = \frac{v_0 \sin \theta}{g}
\]

### b) Horizontal Distance \( d \) from the Building Base to Reach Maximum Height

To find the horizontal distance \( d \) from the base of the building where the water reaches its maximum height, we need to consider the horizontal component of the motion. The horizontal component of the initial velocity \( v_{0x} \) is given by:

\[
v_{0x} = v_0 \cos \theta
\]

Since there is no air resistance, the horizontal velocity remains constant throughout the motion. The horizontal distance \( d \) covered in time \( t_{\text{max}} \) can be expressed as:

\[
d = v_{0x} \cdot t_{\text{max}}
\]

Substitute \( v_{0x} = v_0 \cos \theta \) and \( t_{\text{max}} = \frac{v_0 \sin \theta}{g} \):

\[
d = v_0 \cos \theta \cdot \frac{v_0 \sin \theta}{g}
\]

Simplify the expression:

\[
d = \frac{v_0^2 \sin \theta \cos \theta}{g}
\]

This expression can also be written using the double-angle formula for sine: \( \sin(2\theta) = 2 \sin \theta \cos \theta \):

\[
d = \frac{v_0^2 \sin(2\theta)}{2g}
\]

### Summary

- **a)** The time to reach the maximum height \( t_{\text{max}} \) is:

  \[
  t_{\text{max}} = \frac{v_0 \sin \theta}{g}
  \]

- **b)** The horizontal distance \( d \) from the building base where the water reaches its maximum height is:

  \[
  d = \frac{v_0^2 \sin \theta \cos \theta}{g} \quad \text{or} \quad d = \frac{v_0^2 \sin(2\theta)}{2g}
  \]

Would you like more details about any step or have any questions?

### Related Questions
1. How do you find the maximum vertical height reached by the water?
2. What would be the total time of flight for the water until it reaches the ground again?
3. How would the presence of air resistance affect the time and distance calculated?
4. How can we determine the velocity of the water at a specific point in its trajectory?
5. What happens to the horizontal distance if the angle of projection is increased or decreased?

**Tip:** Understanding the components of projectile motion is essential for analyzing objects moving under the influence of gravity.

Water leaves a fireman’s hose (held near the ground) with an initial velocity v0 = 22.5 m/s at an angle θ = 28.5° above 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. a) Using v0, θ, and g, write an expression for the time, tmax, the water travels to reach its maximum vertical height. b) At what horizontal distance d from the building base, where 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, and g.

Solve a projectile motion problem where water leaves a fireman's hose with an initial velocity of 22.5 m/s at a 28.5° angle. Learn how to calculate the time to reach maximum height and the horizontal distance using key physics principles. Suitable for advanced high school and undergraduate physics students.

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