physics

Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 2.95 m away from a
waterfall 0.54 m in height, at what minimum
speed must a salmon jumping at an angle of
29.7

leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s
2
.
Answer in units of m/s.

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  1. u = V cos 29.7 the whole time
    x = u t
    Vi = V sin 29.7 up velocity at start
    v = Vi - 9.81 t
    h = Vi t - (9.81/2) t^2

    at top, v =0
    so
    t = Vi/9.81
    h = .54 = Vi (Vi/9.81) - (9.81/2)( Vi^2/9.81^2)
    .54 = (1/2)Vi^2/9,81
    1.08*9.81 = Vi^2
    etc
    get Vi
    get t
    get u =2.95/t
    get V = u^2+Vi^2

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    Damon
  2. ah
    V = sqrt (u^2 + Vi^2)

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    Damon
  3. sorry damon but what is the variable u is in this example

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  4. u = horizontal speed the whole time (no horizontal force so no horizontal acceleration

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    Damon
  5. starting from 2.95 m downstream, a line at an angle of 29.7º
    ... passes over the waterfall 1.68 m above the stream

    this means the salmon's trajectory can drop (1.68 - .54) m during the jump

    1.14 = 1/2 g t^2 ... 2.28 / g = t^2 ... t = .48 s

    the horizontal component of the launch velocity is ... 2.95 m / .48s

    the launch velocity is ... (horizontal component) / cos(29.7º)

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  6. I misread it. R Scott did it right.

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    Damon

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