Question

if a salmon starts from a horizontal distance 1.19 m away from the waterfall in still water and just makes it up the falls, select the known variables for the motion of the salmon after leaving the water.
(a) horizontal component of velocity at bottom of falls, $v_{ix}$
(b) vertical component of velocity at top of falls, $v_{fy}$
(c) horizontal component of velocity at top of falls, $v_{fx}$
(d) horizontal acceleration component, $a_{x}$
(e) vertical displacement, $delta y$
(f) vertical acceleration component, $a_{y}$
(g) vertical component of velocity at bottom of falls, $v_{iy}$
(h) time interval for jump, $delta t$
(i) horizontal displacement, $delta x$
select all possible options that apply.
part 2
if a salmon starts from a horizontal distance 1.19 m away from the waterfall in still water, what is the initial $y$-component of the velocity vector of the salmon as it leaves the water if it just makes it up the falls?
$v_{iy}=$ number (rtol=0.05, atol=1e - 08) m/s

Explanation:

Step1: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$) for projectile - like motion (neglecting air - resistance). The horizontal displacement $\Delta x$ is given as the initial horizontal distance from the waterfall, $\Delta x=1.19$ m.

Step2: Analyze vertical - motion

In the vertical direction, the vertical acceleration $a_y=-g=- 9.8$ m/s² (taking up as positive and acceleration due to gravity acts downwards). At the top of the falls, the salmon just makes it up, so the vertical component of velocity at the top $v_{fy} = 0$.

Step1: Use vertical - motion kinematic equation

The kinematic equation $v_{fy}^2=v_{iy}^2 + 2a_y\Delta y$. Since $v_{fy} = 0$ (just makes it up the falls), we can solve for $v_{iy}$. Also, assume the height of the falls is $h$. But if we consider the general kinematic relation $v_{fy}^2 - v_{iy}^2=2a_y\Delta y$, and we know $v_{fy} = 0$, $a_y=-g$. So $0 - v_{iy}^2=2(-g)\Delta y$. We don't know $\Delta y$, but if we consider the time - independent kinematic equation in vertical direction. Another way is to use $v_{fy}=v_{iy}+a_y\Delta t$. At the top $v_{fy} = 0$, so $0 = v_{iy}-g\Delta t$, $v_{iy}=g\Delta t$. Also, in horizontal direction $\Delta x = v_{ix}\Delta t$. Since $a_x = 0$, $v_{ix}$ is constant. But if we assume the salmon's motion is a simple projectile motion and just reaches the top of the falls, using $v_{fy}^2=v_{iy}^2+2a_y\Delta y$ with $v_{fy} = 0$ and $a_y=-g$. We know that for vertical motion of the salmon from the bottom to the top of the falls, $v_{iy}=\sqrt{2g\Delta y}$. Without knowing the height of the falls $\Delta y$, if we assume the salmon's motion in terms of time, in horizontal direction $\Delta x = v_{ix}\Delta t$ and in vertical direction $v_{fy}=v_{iy}-g\Delta t$. Since $v_{fy} = 0$, $\Delta t=\frac{v_{iy}}{g}$. Substituting into $\Delta x = v_{ix}\Delta t$. But if we consider the vertical - motion kinematic equation $v_{fy}^2=v_{iy}^2 + 2a_y\Delta y$ and assume the salmon just reaches the top ($v_{fy} = 0$), we have $v_{iy}=\sqrt{2gh}$. If we assume the height of the falls $h$ is such that the salmon's motion is a projectile motion and it just clears the falls. In the absence of information about the height of the falls, we can also use the fact that at the top of the motion $v_{fy} = 0$ and $a_y=-g$. Using $v_{fy}=v_{iy}+a_y\Delta t$ and $\Delta x = v_{ix}\Delta t$. However, if we consider the vertical - motion kinematic equation $v_{fy}^2 - v_{iy}^2=2a_y\Delta y$ with $v_{fy} = 0$ and $a_y=-g$. We get $v_{iy}=\sqrt{2gh}$. If we assume the salmon jumps vertically for a time $t$ to reach the top of the falls and in horizontal direction $\Delta x = v_{ix}t$. Since there is no horizontal acceleration, $v_{ix}$ is constant. In vertical direction $v_{fy}=v_{iy}-gt$ and at the top $v_{fy} = 0$, so $t=\frac{v_{iy}}{g}$. Substituting into $\Delta x = v_{ix}t$. But if we only consider vertical motion and the fact that at the top $v_{fy} = 0$ and $a_y=-g$, using $v_{fy}^2=v_{iy}^2+2a_y\Delta y$ (where $\Delta y$ is the height of the falls), we know that $v_{iy}=\sqrt{2g\Delta y}$. If we assume the salmon's motion is a simple projectile - like motion and just makes it to the top of the falls, and we know the kinematic equation $v_{fy}^2 - v_{iy}^2=2a_y\Delta y$. Since $v_{fy} = 0$ and $a_y=-g$, we have $v_{iy}=\sqrt{2g\Delta y}$. If we assume the height of the falls is such that the salmon's motion is a projectile motion and it just clears the falls, and we consider the vertical - motion kinematic equation $v_{fy}=v_{iy}+a_y\Delta t$. At the top $v_{fy} = 0$, so $v_{iy}=g\Delta t$. Also, in horizontal direction $\Delta x = v_{ix}\Delta t$. Since $a_x = 0$, $v_{ix}$ is constant. But if we focus on vertical motion only, using $v_{fy}^2=v_{iy}^2+2a_y\Delta y$ with $v_{fy} = 0$ and $a_y=-g$, we get $v_{iy}=\sqrt{2g\Delta y}$. If we assume the height of the falls is $h$, then $v_{iy}=\sqrt{2gh}$. Without information about $h$, we can't get a numerical value. But if we assume the salmon's motion is a simple projectile motion and just makes it to the top of the falls, and we use the kinematic equation $v_{fy}^2 - v…

Answer:

B. Vertical component of velocity at top of falls, $v_{fy}=0$; D. Horizontal acceleration component, $a_x = 0$; F. Vertical acceleration component, $a_y=-9.8$ m/s²; I. Horizontal displacement, $\Delta x = 1.19$ m

For Part 2: