The problem corresponds to the motion of a projectile (the salmon), with initial speed [tex]v_0[/tex], initial direction [tex]\theta=37.6^{\circ}[/tex] and vertical acceleration [tex]g=9.81 m/s^2[/tex] downward. The two equations which gives the horizontal and vertical position of the salmon at time t are
[tex]S_x (t) = v_0 cos \theta t[/tex] (1)
[tex]S_y (t) = v_0 sin \theta t - \frac{1}{2}gt^2[/tex] (2)
We can solve the problem by requiring Sx=3.16 m and Sy=0.379 m, the data of the problem.
Solving eq.(1) for t:
[tex]t=\frac{S_x}{v_0 cos \theta}[/tex]
And substituting this expression of t into eq.(2), we get the following expression for [tex]v_0[/tex]:
[tex]v_0 =\sqrt{\frac{g S_x^2}{2(tan \theta S_x -S_y)cos^2 \alpha}}[/tex]
And substituting the numbers into the equation, we find
[tex]v_0 = 6.16 m/s[/tex]
The minimum speed Salmon must jump is [tex]\boxed{6.16{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}[/tex].
Further explain:
Here, the given problem is on the basis of the projectile motion.
Given:
Initial direction [tex]\left( \theta\right)[/tex] with horizontal is [tex]37.6^\circ[/tex]. Â
The distance travel by Salmon in x-direction at time [tex]t[/tex] is [tex]3.16{\text{ m}}[/tex].
The distance travel by Salmon in y-direction at time [tex]t[/tex] is [tex]0.379{\text{ m}}[/tex]. Â
Acceleration due to gravity is [tex]9.8{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}[/tex]. Â
Formula and concept used:
The velocity component in x-direction [tex]\left( {{v_x}} \right)[/tex] is[tex]v\cos \theta[/tex]. Â
Distance travel by Salmon in x-direction at time [tex]t[/tex] is,
[tex]\begin{gathered}{s_x} = {v_x}t \hfill \\{s_x} = v\cos \theta\cdot t \hfill \\ \end{gathered}[/tex]
Simplify above equation,
[tex]\boxed{t=\dfrac{{{s_x}}}{{v\cos \theta }}}[/tex]              …… (1)
The velocity component in y-direction[tex]\left( {{v_y}} \right)[/tex] is [tex]v\sin \theta[/tex]. Â
Distance travel by Salmon in y-direction at time [tex]t[/tex] is,
[tex]{s_y} = {v_y}t - \frac{1}{2}g{t^2}[/tex]
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Substitute the value of [tex]{v_y}[/tex] in above equation,
[tex]\boxed{{s_y} = v\sin \theta\cdot t - \frac{1}{2}g{t^2}}[/tex]            …… (2)
Substitute the value of [tex]t[/tex] from equation (1) in to the equation (2).
[tex]\begin{aligned}{s_y}&=v\sin \theta\cdot \left( {\frac{{{s_x}}}{{v\cos \theta }}} \right) - \frac{1}{2}g{\left( {\frac{{{s_x}}}{{v\cos \theta }}} \right)^2} \hfill \\&={s_x}\tan\theta- \frac{{g{{\left( {{s_x}} \right)}^2}}}{{2{v^2}{{\cos }^2}\theta}}\hfill\\\end{aligned}[/tex]
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Simplify the above equation to obtain the value of [tex]v[/tex] ,
[tex]\dfrac{{g{{\left( {{s_x}} \right)}^2}}}{{2{v^2}{{\cos }^2}\theta }}={s_x}\tan \theta-{s_y}[/tex]
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Rearrange the above expression.
[tex]\dfrac{1}{{{v^2}}}=\dfrac{{2{{\cos }^2}\theta\cdot \left( {{s_x}\tan\theta- {s_y}} \right)}}{{g{{\left( {{s_x}} \right)}^2}}}[/tex]
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Simplify the above equation,
[tex]{v^2} = \dfrac{{g{{\left( {{s_x}} \right)}^2}}}{{2{{\cos }^2}\theta\cdot \left( {{s_x}\tan \theta- {s_y}} \right)}}[/tex]
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Taking the square root both the sides in above equation,
[tex]\boxed{v=\sqrt {\frac{{g{{\left( {{s_x}} \right)}^2}}}{{2{{\cos }^2}\theta\cdot \left( {{s_x}\tan \theta- {s_y}} \right)}}} }[/tex]                          …… (3)
Calculation:
Substitute the value of [tex]{s_x}[/tex] as [tex]3.16{\text{ m}}[/tex] , the value of [tex]{s_y}[/tex] as [tex]0.379{\text{ m}}[/tex] , the value of [tex]\theta[/tex] Â as [tex]37.6^\circ[/tex] Â in equation (3).
[tex]\begin{aligned}v&=\sqrt {\frac{{9.8{{\left( {3.16} \right)}^2}}}{{2{{\cos }^2}37.6^\circ\cdot\left( {3.16\tan 37.6^\circ- 0.379} \right)}}}\\&=\sqrt {\frac{{97.86}}{{2.579}}}\\&=\sqrt {37.945}\\&=6.16{\text{ }}{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}[/tex] Â
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Learn more:
1. Wind and solar energy: https://brainly.com/question/1062501
2. The threshold frequency of cesium: https://brainly.com/question/6953278
3. Motion under friction https://brainly.com/question/7031524.
Answer detail:
Grade: Senior School
Subject: Physics
Chapter: Projectile motion
Keywords:
Salmon, jump, waterfall, angle of inclination, continue upstream, projectile, acceleration, velocity, acceleration due to gravity, Newton’s laws, equation of motion, velocity, component of acceleration.
