Question

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

Answer 1

Answer:

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed $u$ at an angle of $\theta=38^{\circ}$ 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

$v_x = u cos \theta$

So that the distance travelled is

$d=v_x t = u cos \theta t$ (1)

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

$y = u sin \theta t - \frac{1}{2}gt^2$ (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:

$t=\frac{d}{u cos \theta}$

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

$y = u sin \theta \frac{d}{u cos \theta} - \frac{1}{2}gt^2 = d tan \theta - \frac{1}{2}gt^2\\t = \sqrt{\frac{2(d tan \theta - y)}{g}}=\sqrt{\frac{2(2.33 tan 38^{\circ}-0.488)}{9.81}}=0.521 s$

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

$u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s$