Question

A fire hose nozzle directs a steady stream of water of velocity 50 m/s at an angle of 45° above the horizontal. The stream rises initially but then eventually falls to the ground. Assume water as incompressible and inviscid. Consider the density of air and the air friction as negligible, and assume the acceleration due to gravity as 9.81 m/s\(^2\). The maximum height (in m, round off to two decimal places) reached by the stream above the hose nozzle will then be \(\underline{\hspace{1cm}}\).

Hint:

The maximum height of a projectile is calculated using the vertical component of velocity and the kinematic equation.

Answer 1

Correct Answer: 63.5

The maximum height reached by the stream can be calculated using the kinematic equation for projectile motion: \[ v_y^2 = v_{y0}^2 - 2 g h \] Where: - \( v_y = 0 \) m/s is the final vertical velocity at the maximum height, - \( v_{y0} = 50 \times \sin(45^\circ) = 50 \times \frac{\sqrt{2}}{2} = 35.36 \, \text{m/s} \) is the initial vertical velocity, - \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, - \( h \) is the maximum height. Substituting the values: \[ 0 = (35.36)^2 - 2 \times 9.81 \times h \] \[ h = \frac{(35.36)^2}{2 \times 9.81} = \frac{1250.4}{19.62} \approx 63.9 \, \text{m} \] Thus, the maximum height reached by the stream is \( \boxed{63.90} \, \text{m} \).