A community for students.
Here's the question you clicked on:
 0 viewing
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.)
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.)

This Question is Open

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0yes your fish can do it but you have to post your efforts here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 yvelocity should be 0, giving a final velocity of vf = <6.26cos(θ) m/s, 0.00 m/s>. Now plug values for the ycomponents 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>.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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 = nonreal 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 \(t\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ooooh... 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0great! Note that calculating \(v_0\) was necessary to ensure that the magnitude of the initial velocity is lower than \(6.26m/s\) good work!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.