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 + (1)/(2) at^2`

`-0.449 = 0.576 * v *t - (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 = (2.57)/(0.63)`

`v = 4.1 m/s`

so his minimum speed of jump is 4.1 m/s