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Sagot :
With the projectile launch we can find that the response for the initial velocity of the salmon is 3.68 m/s
Projectile launching is an application of kinematics for movement in two dimensions. In this case they indicate the jump distance x = 2.61 m and the jump height y = 0.277 m and the angle of 39.3º and ask the initial velocity.
Let's use trigonometry for the initial velocity.
cos 39.3 = [tex]\frac{v_x}{v}[/tex]
sin 39.3 = [tex]\frac{v_y}{v}[/tex]
vₓ = v cos 39.3
= v sin 39.3
Suppose that when you reach the height of the waterfall you are in the highest part of the jump, therefore the vertical velocity is zero.
[tex]v_y^2 = v_{oy}^2 -2 g y[/tex]
0 = - 2 g y
[tex]v_{oy}[/tex] = [tex]\sqrt{2gy}[/tex]
Let's calculate
[tex]v_{oy} = \sqrt{2 \ 9.8 \ 0.277}[/tex]
= 2.33 m / s
The relationship of the initial velocity and the angle.
= v sin 39.3
v = [tex]\frac{ v_{oy}}{son 39.3}[/tex]
v = 2.33 / sin 39.3
v = 3.68 m / s
In conclusion using projectile launch we can find the initial velocity of the salmon is 3.68 m / s.
Learn more here: brainly.com/question/15063198
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