Salmon often jump waterfalls to reach their breeding grounds.
starting downstream, 2.9 m away from a waterfall 0.436 m in height, at what minimum speed must a salmon jumping at an angle of 44.7◦ leave the water to continue upstream? the acceleration due to gravity is 9.81 m/s2 .

5.79 m/s

Explanation:

R=2.9 m

H=0.436 m

∅=44.7°

For Horizontal Motion

aₓ=0      uₓ=ucos∅

x=uₓt+

R=ucos∅*t+0

2.9=ucos(38.3)*t

t=

For Vertical Motion

Y=Uy*t+

By putting value

0.436=usin∅()-()

0.436=sin(44.7)()-()

0.436=2.869-

u²=33.54

u=5.79 m/s

5.83 m/s

Explanation:

This is a projectile motion problem.

Let's call

the initial velocity

In x-coordinate:

where α is 44.7°

In y-coordinate:

In x-coordinate the displacement is:

where t is time. Isolating the initial speed and replacing with x = 2.9 (the distance travelled in x direction) :

In y-coordinate the displacement is:

Replacing with the initial velocity, y = 0.436, alpha = 44.7° and g = 9.81:

Replacing this value in the previous initial velocity equation:

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray