Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
