Question

Salmon often jump waterfalls to reach their breeding grounds. Starting 3.15 m from a waterfall 0.45 m in height, at what minimum speed must a salmon jumping at an angle of 33.2 degree leave the water to continue upstream? The acceleration due to gravity is 9.81 m/s2. Answer in units of m/s.

Answer 1

Let the salmon jump with speed v at an angle of 33.3 degree

So here we can find out the components of his velocity in x and y direction

`v_x = v cos33.2`

`v_y = v sin33.2`

now the horizontal displacement of the salmon is 3.15 m so he cover this horizontal distance with constant speed as there is no acceleration in x direction

here we can say

`v_x * t = 3.15`

`v cos33.2 * t = 3.15`

`v*t = 3.76`

now in Y direction it is an accelerated motion as it is accelerated due to gravity

`y = v_y * t + (1)/(2)at^2`

`-0.45 = v sin33.2* t - (1)/(2)*9.8* t^2`

now we will plug in v*t = 3.76

`-0.45 = 3.76 * sin33.2 - 4.9 * t^2`

`4.9 t^2 = 2.51`

`t^2 = 0.512`

`t = 0.72 s`

now speed v is given by equation above

`v*t = 3.76`

`v*0.72 = 3.76`

`v = 5.25 m/s`

so here he must have to jump with minimum speed of 5.25 m/s