Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over waterfalls as high as 2.83 m. With what minimum speed (in m/s) must a salmon launch itself into the air to clear a 2.83-m waterfall

minimum speed of launch must be 7.45m/s

Explanation:

Given the following:

Height or distance (s) = 2.83m

The final velocity(Vf) at maximum height = 0

Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2

From the 3rd equation of motion:

V^2 = u^2 - 2gs

Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

u = √0^2 - 2(-9.8)(2.83)

u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

(75°f − 32) × 5/9 = 23.889°c