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Sav [38]
3 years ago
5

Salmon often jump waterfalls to reach their

Physics
1 answer:
natta225 [31]3 years ago
5 0

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

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

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}

and we can solve this formula for u, the initial speed of the salmon:

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

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