Answer:
5.93 m/s (2 d.p.)
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:
- [tex]\textsf{Horizontal component of $u= u \cos \theta$}[/tex]
- [tex]\textsf{Vertical component of $u= u \sin\theta$}[/tex]
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)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{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}}[/tex]
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:
- [tex]\textsf{Horizontal component of $u= u \cos 36.2^{\circ}$}[/tex]
- [tex]\textsf{Vertical component of $u= u \sin36.2^{\circ}$}[/tex]
Resolving horizontally
The horizontal component of velocity is constant, as there is no acceleration horizontally.
Resolving horizontally, taking → as positive:
[tex]s=3.02 \quad u=u \cos 36.2^{\circ} \quad v=u \cos 36.2^{\circ} \quad a=0[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & =ut+\dfrac{1}{2}at^2\\\\3.02 & = (u \cos 36.2^{\circ})t+\dfrac{1}{2}(0)t^2\\3.02 & = (u \cos 36.2^{\circ})t\\\implies t&=\dfrac{3.02}{u \cos 36.2^{\circ}}\end{aligned}[/tex]
Resolving vertically
Acceleration due to gravity = 9.81 ms⁻²
Resolving vertically, taking ↑ as positive and using the found expression for t:
[tex]s=0.258 \quad u=u \sin 36.2^{\circ} \quad a=-9.81 \quad t=\dfrac{3.02}{u \cos 36.2^{\circ}}[/tex]
[tex]\begin{aligned}\textsf{Using} \quad s & =ut+\dfrac{1}{2}at^2\\\\0.258 & = (u \sin 36.2^{\circ})\left(\dfrac{3.02}{u \cos 36.2^{\circ}}\right)+\dfrac{1}{2}(-9.81)\left(\dfrac{3.02}{u \cos 36.2^{\circ}}\right)^2\\0.258&=3.02 \tan36.2^{\circ}-4.905\left(\dfrac{9.1204}{u^2 \cos^2 36.2^{\circ}}\right)\\0.258-3.02 \tan36.2^{\circ}&=-\dfrac{44.735562}{u^2 \cos^2 36.2^{\circ}}\\u^2&=-\dfrac{44.735562}{(0.258-3.02 \tan36.2^{\circ})(\cos^2 36.2^{\circ})}\\u^2&=35.18849443\\ u&=5.931989079\end{aligned}[/tex]
Therefore, the minimum speed at which the salmon should leave the water is 5.93 m/s (2 d.p.).
