Question

Salmon often jump waterfalls to reach their

breeding grounds.
Starting downstream, 3.02 m away from a
waterfall 0.258 m in height, at what minimum
speed must a salmon jumping at an angle of
36.2° leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s².
Answer in units of m/s.

Answer 1

easeaseaseas

Step-by-step explanation:

Answer 2

5.93 m/s (2 d.p.)

Step-by-step explanation:

When a body is projected through the air with initial speed u, at an angle of θ to the horizontal, it will move along a curved path.

Therefore, trigonometry can be used to resolve the body's initial velocity into its vertical and horizontal components:

• 
  `\textsf{Horizontal component of $u= u \cos \theta$}`
• 
  `\textsf{Vertical component of $u= u \sin\theta$}`

As the projectile is modeled as moving only under the influence of gravity, the only acceleration the projectile will experience will be acceleration due to gravity.

Constant Acceleration Equations (SUVAT)

`\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+(1)/(2)at^2\\\\ s&=\left((u+v)/(2)\right)t\\\\v^2&=u^2+2as\\\\s&=vt-(1)/(2)at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^(-1)$\\\\$v$ = final velocity in ms$^(-1)$\\\\$a$ = acceleration in ms$^(-2)$\\\\$t$ = time in s (seconds)\end{minipage}}`

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

If the salmon jumps at an angle of 36.2° then:

• 
  `\textsf{Horizontal component of $u= u \cos 36.2^(\circ)$}`
• 
  `\textsf{Vertical component of $u= u \sin36.2^(\circ)$}`

Resolving horizontally

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

`s=3.02 \quad u=u \cos 36.2^(\circ) \quad v=u \cos 36.2^(\circ) \quad a=0`

`\begin{aligned}\textsf{Using} \quad s & =ut+(1)/(2)at^2\\\\3.02 & = (u \cos 36.2^(\circ))t+(1)/(2)(0)t^2\\3.02 & = (u \cos 36.2^(\circ))t\\\implies t&=(3.02)/(u \cos 36.2^(\circ))\end{aligned}`

Resolving vertically

Acceleration due to gravity = 9.81 ms⁻²

Resolving vertically, taking ↑ as positive and using the found expression for t:

`s=0.258 \quad u=u \sin 36.2^(\circ) \quad a=-9.81 \quad t=(3.02)/(u \cos 36.2^(\circ))`

`\begin{aligned}\textsf{Using} \quad s & =ut+(1)/(2)at^2\\\\0.258 & = (u \sin 36.2^(\circ))\left((3.02)/(u \cos 36.2^(\circ))\right)+(1)/(2)(-9.81)\left((3.02)/(u \cos 36.2^(\circ))\right)^2\\0.258&=3.02 \tan36.2^(\circ)-4.905\left((9.1204)/(u^2 \cos^2 36.2^(\circ))\right)\\0.258-3.02 \tan36.2^(\circ)&=-(44.735562)/(u^2 \cos^2 36.2^(\circ))\\u^2&=-(44.735562)/((0.258-3.02 \tan36.2^(\circ))(\cos^2 36.2^(\circ)))\\u^2&=35.18849443\\ u&=5.931989079\end{aligned}`

Therefore, the minimum speed at which the salmon should leave the water is 5.93 m/s (2 d.p.).