Solution 18
1. Mass Flow Rate
The rate at which mass leaves the hose is given by: $$ \frac{dm}{dt} = \rho A u $$ where $\rho$ is the density of water, $A$ is the cross-sectional area, and $u$ is the exit velocity.
The rate at which mass leaves the hose is given by: $$ \frac{dm}{dt} = \rho A u $$ where $\rho$ is the density of water, $A$ is the cross-sectional area, and $u$ is the exit velocity.
2. Time of Flight ($T$)
Since the water stream is continuous, the total mass in the air at any instant is equal to the mass flow rate multiplied by the time a single drop stays in the air. Using the equation of motion for vertical displacement (taking upward as positive and origin at the nozzle): $$ y(t) = u_y t – \frac{1}{2}gt^2 $$ Here, the water hits the ground at displacement $y = -h$. The initial vertical velocity is $u_y = u \sin\theta$. $$ -h = (u \sin\theta)T – \frac{1}{2}gT^2 $$ Rearranging: $$ \frac{1}{2}gT^2 – (u \sin\theta)T – h = 0 $$ Solving for $T$ using the quadratic formula: $$ T = \frac{u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh}}{g} $$ (We take the positive root for time).
Since the water stream is continuous, the total mass in the air at any instant is equal to the mass flow rate multiplied by the time a single drop stays in the air. Using the equation of motion for vertical displacement (taking upward as positive and origin at the nozzle): $$ y(t) = u_y t – \frac{1}{2}gt^2 $$ Here, the water hits the ground at displacement $y = -h$. The initial vertical velocity is $u_y = u \sin\theta$. $$ -h = (u \sin\theta)T – \frac{1}{2}gT^2 $$ Rearranging: $$ \frac{1}{2}gT^2 – (u \sin\theta)T – h = 0 $$ Solving for $T$ using the quadratic formula: $$ T = \frac{u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh}}{g} $$ (We take the positive root for time).
3. Total Mass Calculation
The total mass $M$ is: $$ M = \frac{dm}{dt} \times T = \rho A u \left( \frac{u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh}}{g} \right) $$
The total mass $M$ is: $$ M = \frac{dm}{dt} \times T = \rho A u \left( \frac{u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh}}{g} \right) $$
4. Numerical Substitution
Given: $h = 0.8$ m, $u = 6$ m/s, $\theta = 30^\circ$, $g = 10$ m/s$^2$. $$ u \sin 30^\circ = 6(0.5) = 3 \text{ m/s} $$ $$ T = \frac{3 + \sqrt{3^2 + 2(10)(0.8)}}{10} = \frac{3 + \sqrt{9 + 16}}{10} = \frac{3 + 5}{10} = 0.8 \text{ s} $$ Using the values from the provided answer key context (Answer = 72g), we can infer the Area $A$ was likely meant to be consistent with typical problem values (likely in cm$^2$ or mm$^2$). However, the algebraic solution holds: $$ M = \rho A u (0.8) = 1000 \times A \times 6 \times 0.8 = 4800 A \, \text{kg} $$ *(Note: If the answer is 72g = 0.072 kg, then $A = 0.15 \text{ cm}^2$)*
Given: $h = 0.8$ m, $u = 6$ m/s, $\theta = 30^\circ$, $g = 10$ m/s$^2$. $$ u \sin 30^\circ = 6(0.5) = 3 \text{ m/s} $$ $$ T = \frac{3 + \sqrt{3^2 + 2(10)(0.8)}}{10} = \frac{3 + \sqrt{9 + 16}}{10} = \frac{3 + 5}{10} = 0.8 \text{ s} $$ Using the values from the provided answer key context (Answer = 72g), we can infer the Area $A$ was likely meant to be consistent with typical problem values (likely in cm$^2$ or mm$^2$). However, the algebraic solution holds: $$ M = \rho A u (0.8) = 1000 \times A \times 6 \times 0.8 = 4800 A \, \text{kg} $$ *(Note: If the answer is 72g = 0.072 kg, then $A = 0.15 \text{ cm}^2$)*
General Result:
$$ M = \frac{\rho A u}{g} \left( u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh} \right) $$
$$ M = \frac{\rho A u}{g} \left( u \sin\theta + \sqrt{u^2 \sin^2\theta + 2gh} \right) $$