Question

Salmon often jump waterfalls to reach their breeding grounds.

Starting downstream, 2.9 m away from a waterfall 0.436 m in height, at what minimum speed must a salmon jumping at an angle of 44.7◦ leave the water to continue upstream? The acceleration due to gravity is 9.81 m/s2 .

Answer 1

5.83 m/s

Step-by-step explanation:

This is a projectile motion problem.

Let's call

`V_0` the initial velocity

In x-coordinate:

`V_(0x) = V_0 * cos \alpha`

where α is 44.7°

In y-coordinate:

`V_(0y) = V_0 * sin \alpha`

In x-coordinate the displacement is:

`x = V_0 * cos \alpha * t`

where t is time. Isolating the initial speed and replacing with x = 2.9 (the distance travelled in x direction) :

`V_0 = (2.9)/(cos \alpha * t)`

In y-coordinate the displacement is:

`y = V_0 * sin \alpha * t - 1/2 * g * t^2`

Replacing with the initial velocity, y = 0.436, alpha = 44.7° and g = 9.81:

`0.436 = (2.9)/(cos \alpha * t) * sin \alpha * t - 1/2 * g * t^2`

`0.436 =2.9 * tan 44.7 - 1/2 * 9.81 * t^2`

`t^2 = (2.9 * tan 44.7 - 0.436)/( 1/2 * 9.81)`

`t = √(0.495)`

`t = 0.7`

Replacing this value in the previous initial velocity equation:

`V_0 = (2.9)/(cos 44.7 * 0.7)`

`V_0 = 5.83 \; m/s`

Answer 2

5.79 m/s

Step-by-step explanation:

R=2.9 m

H=0.436 m

∅=44.7°

For Horizontal Motion

aₓ=0 uₓ=ucos∅

x=uₓt+
`(1axt^(2) )/(2)`

R=ucos∅*t+0

2.9=ucos(38.3)*t

t=
`(4.079)/(u)`

For Vertical Motion

Y=Uy*t+
`(1ayt^(2) )/(2)`

By putting value

0.436=usin∅(
`(4.079)/(u)`)-(
`(1)/(2)g *((4.079)/(u) )^(2)`)

0.436=sin(44.7)(
`\frac{4.079}`)-(
`(1)/(2)g *((4.079)/(u) )^(2)`)

0.436=2.869-
`(81.61)/(u^(2) )`

u²=33.54

u=5.79 m/s