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anonymous

  • one year ago

Suppose a chinook salmon needs to jump a waterfall that is 1.50 m high. If the fish starts from a distance 1.00 m from the base of the ledge over which the waterfall flows, find the x- and y- components of the initial velocity the salmon would need to just reach the ledge at the top of its trajectory. Can the fish make this jump? (A chinook salmon can jump out of the water with a speed of 6.26 m/s.)

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  1. IrishBoy123
    • one year ago
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    yes your fish can do it but you have to post your efforts here.

  2. anonymous
    • one year ago
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    I took up like three pages of a notebook trying to figure it out over and over... The fish can make the jump, but it never actually satisfies the requirement that it reaches the ledge at the top of its trajectory. I ended up just making a list of the defining equations of... acceleration: \[a_{x}=0.00 \frac{m}{s^2},a_{y}=-9.80 \frac{m}{s^2}\] Velocity: \[v_{x}=6.26 cos(\theta) \frac{m}{s}, v_{y}=6.26 sin(\theta) \frac{m}{s} - 9.80t \frac{m}{s^2}\] and position: \[r_{x}=6.26 cos(\theta) t \frac{m}{s}, r_{y}=6.26 sin(\theta) t \frac{m}{s} - 4.90t^2 \frac{m}{s^2}\] Use r0 = <0.00 m, 0.00 m>, rf = <1.00 m, 1.50 m> for initial and final positions. Use v0 = <6.26cos(θ) m/s, 6.26sin(θ) m/s> for the initial velocity, and since the landing needs to be at the top of the trajectory, the final y-velocity should be 0, giving a final velocity of vf = <6.26cos(θ) m/s, 0.00 m/s>. Now plug values for the y-components into v^2 = v0^2 + 2aΔy... (0.00 m/s)^2 = (6.26sin(θ) m/s)^2 - 2(9.80 m/s^2)(1.50 m), which reduces to sin^2(θ) = .750, giving θ = +/- 60.0 degrees. Disregard the negative value, because we know the salmon has to jump upward for it to reach the ledge. The angle has to be 60.0 degrees. This makes v0 = <3.13 m/s, 5.42 m/s>.

  3. anonymous
    • one year ago
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    But it produced inconsistencies... Find the time it takes, for instance, using this initial velocity. First: rfx = r0x + v0x*t => 1.00 m = 0.00 m + (3.13 m/s)*t ==> t = .319 s Now: rfy = r0y + v0y*t - g/2 * t^2 => 1.50 m = 0.00 m + (5.42 m/s)*t - (4.90 m/s^2)*t^2 => t = (-5.42 +/- sqrt(5.42^2 - 4*-4.90*-1.50))/(2*-4.90) s = non-real answer??? Well, try something else: vfy = v0y - gt => 0 m/s = 5.42 m/s - (9.80 m/s^2)*t => t = .553 s, which isn't equal to the first value for t anyway. So... it's just confusing me way more than it probably should.

  4. anonymous
    • one year ago
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    I think the hint : "A chinook salmon can jump out of the water with a speed of 6.26 m/s" is giving you the maximum possible velocity. i would solve for the initial velocity components \(v_{0x}\) and \(v_{0y}\), while ensuring the final position and final velocity you said, \(\vec r_f = <1, 1.5>\) and \(\vec v_f = <v_{0x}, 0>\), that way we have, evaluating everything at the final time: \(v_y = 0 = v_{0y}-gt\) \(y = 1.5 = v_{0y}t-\frac{1}{2}gt^2\) \(x = 1 = v_{0x}t\) and we need to solve for \(v_{0x}\), \(v_{0y}\) and \(t\)

  5. anonymous
    • one year ago
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    Ooooh... okay. I didn't think of treating the magnitude of the initial velocity as variable with a maximum of 6.26 m/s. This makes much more sense now. I solved it and got the values \[t=.533 s, v_{0y}=5.42 m/s, v_{0x}=1.88 m/s\] and by extension (though not really necessary) \[v_{0} = 5.74 m/s, \theta=70.9^{\circ}.\] Thank you! That cleared up a lot of confusion I was having.

  6. anonymous
    • one year ago
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    great! Note that calculating \(v_0\) was necessary to ensure that the magnitude of the initial velocity is lower than \(6.26m/s\) good work!

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