Salmon often jump waterfalls to reach their
breeding grounds.
starting downstream, 2.15 m away from a
waterfall 0.587 m in height, at what minimum
speed must a salmon jumping at an angle of
26.5° leave the water to continue upstream?
the acceleration due to gravity is 9.81 m/s.
answer in units of m/s.

Minimum speed with which the salmon must jump at an angle of 26.5° to continue upstream is 7.65 m/s.

Explanation:

The Salmon's jump is following a projectile motion.

A projectile motion is the motion of an object in a curved trajectory near the surface of the Earth when it is thrown with a certain initial velocity at a certain angle with the horizontal, under the action of gravitational force.

In this case, the height of launch is different from the height of landing so we will use equations of motion to solve this problem.

X = range = 2.15 m

h = height = 0.587 m

Initial speed = v = ?

Angle of jump = α = 26.5°

gravitational acceleration = g = 9.81 m/s²

Using the 2nd equation of motion:

h = vsinα*t - (1/2)(g/t²)

0.587 = vsin(26.5°)*t - (1/2)(9.81/t²)

0.587 = vsin(26.5°)*t - 4(1/t²) eq.1

To find 'v', we must find the value of t.

Finding 't':

We know that d = v*t

⇒ X = (Vx) *t

X is horizontal range, so we are using the x-component of v here, t is unknown for now.

⇒ 2.15 = Vcosα * t

⇒ t = 2.15/Vcos(26.5°) eq.2

using the value of 't' from eq.2 in eq.1

0.587 = {vsin(26.5°)/vcos(26.5°)}*(2.15)  -  4.9((2.15)²/v²cos²(26.5°))

0.587 = tan(26.5°)*2.15 - (4.9)*(4.63)/v²(0.8)²

0.587 = 0.498*2.15 - 22.65/v²(0.8)

0.587 = 1.0707 - 28.31/v²

28.31 = 0.4837*v²

Taking under-root:

v = √(28.31/0.4837)

v = √58.52

v = 7.65 m/s

what is the question?