Salmon often jump upstream through waterfalls to reach their breeding grounds. One salmon came across a waterfall 1.30 m in height, which she jumped in 1.3 s at an angle of 72° to continue upstream. What was the initial speed of her jump? (Assume the launch angle is measured above the horizontal.)

Answer 1

x = v(ox)•t =v(o)•cosα•t,
h =v(oy) •t - g•t²/2 = v(o)•sinα•t- g•t²/2,

x = v(o)•cosα•t,
h + g•t²/2 = v(o)•sinα•t,
(2h+ g•t²)/2•x = sinα/cosα = tanα,

x =(2h+ g•t²)/2•tanα =
=(2•1.3 +9.8•1.3²)/2•tan72º= 3.11 m.

v(o) = x/cosα•t =3.11/cos72º•1.3 =7.74 m/s.