Answer:
5.68 m/s
Explanation:
The motion of the salmon is the same as a projectile: it is launched with an initial speed [tex]u[/tex] at an angle of [tex]\theta=38^{\circ}[/tex] 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
[tex]v_x = u cos \theta[/tex]
So that the distance travelled is
[tex]d=v_x t = u cos \theta t[/tex] (1)
- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is
[tex]y = u sin \theta t - \frac{1}{2}gt^2[/tex] (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:
[tex]t=\frac{d}{u cos \theta}[/tex]
And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:
[tex]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[/tex]
And then, we can use eq.(1) again to find the initial speed, u:
[tex]u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s[/tex]
