Answer:
5.0 m/s
Explanation:
The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is
[tex]v_x = u cos \theta[/tex]
where u is the initial speed and [tex]\theta=37.7^{\circ}[/tex]. The horizontal distance travelled by the salmon is
[tex]d=v_x t = (ucos \theta)t[/tex]
where d = 1.95 m and t is the time needed to reach the final point.
Re-arranging for t,
[tex]t=\frac{d}{v_x}=\frac{d}{u cos \theta}[/tex] (1)
Along the vertical direction, the equation of motion is
[tex]y=h+u_y t -\frac{1}{2}gt^2[/tex]
where:
y = 0.311 m is the final height reached by the salmon
h = 0 is the initial height
[tex]u_y = u sin \theta[/tex] is the vertical component of the initial velocity of the salmon
[tex]g=9.81 m/s^2[/tex] is the acceleration of gravity
t is the time
Substituting t as found in eq.(1), we get the equation
[tex]y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}[/tex]
and we can solve this formula for u, the initial speed of the salmon:
[tex]y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s[/tex]
