Question

an oscillating garden sprinkler which discharges water with an initial velocity v_0 of 8 m/s is used to water a vegetable garden. determine the distance d to the farthest point b that will be watered and the corresponding angle a when (a) the vegetables are just beginning to grow, (b) the height h of the corn is 1.8 m.

Explanation:

Step1: Recall projectile - motion equations

The horizontal range formula for projectile motion is $d=\frac{v_{0}^{2}\sin2\alpha}{g}$, where $v_{0}$ is the initial velocity, $\alpha$ is the launch - angle, and $g = 9.8\ m/s^{2}$ is the acceleration due to gravity. The vertical displacement equation is $y=v_{0y}t-\frac{1}{2}gt^{2}$, where $v_{0y}=v_{0}\sin\alpha$ and $t$ is the time of flight.

Step2: Find the maximum range (part a)

The range $d=\frac{v_{0}^{2}\sin2\alpha}{g}$ is maximized when $\sin2\alpha = 1$. Since $\sin2\alpha = 1$ when $2\alpha=90^{\circ}$, then $\alpha = 45^{\circ}$. Given $v_{0}=8\ m/s$ and $g = 9.8\ m/s^{2}$, we substitute into the range formula:
\[d=\frac{v_{0}^{2}\sin2\alpha}{g}=\frac{8^{2}\times1}{9.8}=\frac{64}{9.8}\approx6.53\ m\]

Step3: Solve for the angle in part (b)

The vertical displacement equation is $y = v_{0}\sin\alpha t-\frac{1}{2}gt^{2}$, and the horizontal displacement $x = v_{0}\cos\alpha t$. From $x = v_{0}\cos\alpha t$, we have $t=\frac{x}{v_{0}\cos\alpha}$. Substitute $t$ into the vertical - displacement equation: $y = v_{0}\sin\alpha\frac{x}{v_{0}\cos\alpha}-\frac{1}{2}g(\frac{x}{v_{0}\cos\alpha})^{2}$.
We know $y = 1.8\ m$, $v_{0}=8\ m/s$, and $x$ is the range $x = d=\frac{v_{0}^{2}\sin2\alpha}{g}$.
Another way is to use the kinematic equation $v_{y}^{2}=v_{0y}^{2}-2gy$. First, $v_{0y}=v_{0}\sin\alpha$ and $v_{y}$ at the maximum height of the trajectory for the vertical motion of the water droplets.
We know $v_{y}^{2}=v_{0}^{2}\sin^{2}\alpha - 2gy$. At the end - point of the trajectory in the vertical direction, when the water reaches the height $y = 1.8\ m$.
\[v_{y}^{2}=v_{0}^{2}\sin^{2}\alpha-2gy\]
\[0 = v_{0}^{2}\sin^{2}\alpha-2gy\]
\[\sin^{2}\alpha=\frac{2gy}{v_{0}^{2}}\]
Substitute $y = 1.8\ m$, $v_{0}=8\ m/s$, and $g = 9.8\ m/s^{2}$:
\[\sin^{2}\alpha=\frac{2\times9.8\times1.8}{8^{2}}=\frac{35.28}{64}\]
\[\sin\alpha=\sqrt{\frac{35.28}{64}}\approx0.74\]
\[\alpha=\sin^{- 1}(0.74)\approx47.8^{\circ}\]

Answer:

(a) $d\approx6.53\ m$, $\alpha = 45^{\circ}$
(b) $\alpha\approx47.8^{\circ}$