Question

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

Answer 1

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s