Question

a water fountain ejects water at a 73 degree angle from a height of 10 cm and lands 20 cm away with a final height of 0 cm What is the initial speed of the water? how high does the water go?

Answer 1

Task. A water fountain ejects water at a 73 degree angle from a height of 10 cm and lands 20 cm away with a final height of 0 cm. What is the initial speed of the water? How high does the water go?

Solution. Let $v_{0}$ be the initial velocity of the water and $\phi=73{}^{\circ}$ be the angle of the initial velocity with horizontal axis. Then its motion can be decomposed as a sum of vertical and horizontal motion. The vertical motion has constant acceleration $g=9.88\ m/s^{2}$ due to gravity and initial velocity $v_{y}^{0}=v_{0}\sin\phi$, while the horizontal motion has constant velocity $v_{x}^{0}=v_{0}\cos\phi$.

By assumption the water passes distance $d_{1}=20\ cm$ away from the initial horizontal position before landing the ground, therefore the time of water motion is

$t_{1}=\frac{d_{1}}{v_{x}^{0}}=\frac{d_{1}}{v_{0}\cos\phi}.$

Let $h_{0}=10\ cm$ be the initial height of the water. Then the height at time $t$ is given by the formula:

$h(t)=h_{0}+v_{y}^{0}t-\frac{gt^{2}}{2},$

and we by assumption

$h(t_{1})=0.$

Substituting $t_{1}$ we get:

$=h_{1}(t_{1})=h_{0}+v_{y}^{0}t_{1}-\frac{gt_{1}^{2}}{2}=h_{0}+v_{0}\sin\phi\cdot\frac{d_{1}}{v_{0}\cos\phi}-\frac{g\cdot\left(\frac{d_{1}}{v_{0}\cos\phi}\right)^{2}}{2}$

$=h_{0}+d_{1}\tan\phi-\frac{gd_{1}^{2}}{2v_{0}^{2}\cos^{2}\phi}$

Hence

$v_{0}=\sqrt{\frac{gd_{1}^{2}}{2(h_{0}+d_{1}\tan\phi)\cos^{2}\phi}}=\frac{d_{1}}{\cos\phi}\cdot\sqrt{\frac{g}{2(h_{0}+d_{1}\tan\phi)}}$

Substituting values we get:

$v_{0}=\frac{d_{1}}{\cos\phi}\cdot\sqrt{\frac{g}{2(h_{0}+d_{1}\tan\phi)}}=\frac{0.2}{\cos 73{}^{\circ}}\cdot\sqrt{\frac{9.81}{2\cdot(0.1+0.2\cdot\tan 73{}^{\circ})}}\approx 1.74\ m/s.$

Notice that the vertical velocity of the water is given by the formula:

$v_{y}(t)=v_{y}^{0}-gt.$

The maximal height of water is achieved at time $t$ when $v_{y}(t)=0$:

$v_{y}(t)=v_{y}^{0}-gt=0,\qquad\Rightarrow\qquad t=\frac{v_{y}^{0}}{g}=\frac{v_{0}\sin\phi}{g}.$

Substituting this $t$ into the formula for $h(t)$ we get the maximal values of height:

$h(t)$ $=h_{0}+v_{0}\sin\phi\cdot\frac{v_{0}\sin\phi}{g}-\frac{g\left(\frac{v_{0}\sin\phi}{g}\right)^{2}}{2}$

$=h_{0}+\frac{(v_{0}\sin\phi)^{2}}{g}-\frac{(v_{0}\sin\phi)^{2}}{2g}=h_{0}+\frac{(v_{0}\sin\phi)^{2}}{2g}$

Hence

$\max(h)=h_{0}+\frac{(v_{0}\sin\phi)^{2}}{2g}=0.1+\frac{(1.74\cdot\sin 73{}^{\circ})}{2\cdot 9.81}\approx 24cm.$

Answer. $v_{0}=1.74\ m/s,\quad\max h=24\ cm$.