Salmon often jump waterfalls to reach their breeding grounds. starting downstream, 2.33 m away from a waterfall 0.488 m in height, at what minimum speed must a salmon jumping at an angle of 38 ◦ leave the water to continue upstream? the acceleration due to gravity is 9.81 m/s 2

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed at an angle of above the horizontal.

The motion of the salmon consists of two indipendent motion:

- Along the horizontal direction, it is a uniform motion with constant velocity

So that the distance travelled is

(1)

- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is

(2)

where g is the acceleration of gravity.

We know the following:

- The horizontal distance travelled by the salmon to reach the waterfall is

d = 2.33 m

- The vertical distance travelled is the height of the waterfall,

y = 0.488 m

From (1) we get:

And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:

And then, we can use eq.(1) again to find the initial speed, u:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft