I need help finding out why my solution is not correct! Note question is in the

Question

I need help finding out why my solution is not correct! Note question is in the context of Relative Velocity, as it is from that section of the physics textbook. 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? (Remember that a Chinook salmon can jump out of the water with a speed of 6.26 m/s.) Please state ehy my solution is not right, and when explaining, include a diagram showing your work. Thank you so much for the help, I really appreciate it!

Explanation / Answer

You are given that the fish jumps to a distance of  d = 1.00 m horizontally and    h = 1.50m vertically. . If we use basic kinematics in each direction, we get: .      d = v cos t          and       h = v sin t - (1/2) g t2 . Also, you know that if he barely makes it, his final verticalvelocity is zero, so... .              0 = v sin - g t . You now have three equations and threeunknowns:    v and t.   If we eliminate v nd t, we can solve for: .           vsin = gt             t = d / vcos   .       h = ( gt )t    -    (1/2) g t2 .        h =    (1/2) gt2 .       h =   (1/2) (v sin) (d / vcos)   .      2 h =   dtan .      2 * 1.50 = 1.00tan          =   71.565 . Now use this to find v: .    v sin = g d / vcos .         v2 = g d / sin cos = 9.80 * 1.00 /sin71.565 cos71.565   =   32.6667 .        v =    5.715  m/s     is the minimum initial speed of thesalmon .        h =    (1/2) gt2 .       h =   (1/2) (v sin) (d / vcos)   .      2 h =   dtan .      2 * 1.50 = 1.00tan          =   71.565 . Now use this to find v: .    v sin = g d / vcos .         v2 = g d / sin cos = 9.80 * 1.00 /sin71.565 cos71.565   =   32.6667 .        v =    5.715  m/s     is the minimum initial speed of thesalmon

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