A Chinook salmon can jump out of water with a speed of 6.70 m/s6.70 m/s . How far horizontally dd can a Chinook salmon travel through the air if it leaves the water with an initial angle of θ=29.0°θ=29.0° with respect to the horizontal? (Let the horizontal direction the fish travels be in the +x+x direction, and let the upward vertical direction be the +y+y direction. Neglect any effects due to air resistance.)

R = 3.88 m

Explanation:

As the Chinook salmon leaves the water till it gets back into the water it is performing a projectile motion with the following parameters:

V₀ = Launch Speed = 6.7 m/s

θ = Launch Angle = 29°

R= Range of Projectile= Horizontal Distance Covered by Chinook salmon= ?

The value of the range of a projectile is given by the following formula:

R = (V₀² Sin 2θ)/g

R = [(6.7 m/s)² Sin {(2)(29°)}/(9.8 m/s²)]

R = [(6.7 m/s)² Sin (58°)/(9.8 m/s²)]

R = 3.88 m

thermal energy is the answer