2D Projectile Motion - Minimum Speed

[stickyriceyum]

Homework Statement

Salmon often jum waterfalls to reach their breeding grounds.
The acceleration of gravity is 9.81 m/s^2
Starting 1.73 m from a waterfall .0386 m in height, at what minimum speed must a salmon jumping at an angle of 44.3 degrees leave the water to continue upstream? Answer in units of m/s.

Homework Equations

Vox = Vocos(theta)
Vox = 6.179cos(44.3) = 4.42
Voy = Vosin(theta)
Voy = 6.179sin(44.3) = 4.316

The Attempt at a Solution

- I drew a graph:
x-axis labeled as 1.73m and the y-axis labeled as .386m

- Tried to find time:
t = (sqrt) 2(h)/g
t = (sqrt) 2(.386)/9.81
t = .28 seconds <-- I think this may be wrong.

- Tried to find Vo:
Vo = x/t
Vo = 1.73/.28
Vo = 6.179

Now plugged Vo result to find components:
Vox = Vocos(theta)
Vox = (6.179)cos(44.3)
Vox = 4.42

Voy = Vosin(theta)
Voy = (6.179)sin(44.3)
Voy = 4.316

... I'm not sure what to do next. Am I supposed to use the Pythagorean Theorem? And find the tangent? I have no clue. All help would be appreciated. =)

 

[nasu]

I am sorry to ask this, but are you sure you did actually read the problem?
You have to FIND the initial velocity, it is not give.
The angle is given, so you don't need to find it (by using tangent...).
Where did you get the value of 6.179 from?

"t = (sqrt) 2(h)/g
t = (sqrt) 2(.386)/9.81
t = .28 seconds <-- I think this may be wrong."

You are right (that it is wrong). The vertical motion has initial speed (voy):
y=voy*t-1/2*g*t^2

What yo have to do is to write equations of motion for both x and y direction and solve them to find vox and voy.
Hint: you can start by finding what is voy in order to reach the given height.

 

[baba89]

As a scientist, your approach to solving this problem is correct so far. You have correctly found the horizontal and vertical components of the initial velocity of the salmon.

To find the minimum speed required for the salmon to jump the waterfall, you need to consider the conservation of energy. At the top of the waterfall, the salmon will have both potential and kinetic energy. At the bottom of the waterfall, the salmon will only have kinetic energy. Therefore, the minimum speed required for the salmon to jump the waterfall will be equal to the initial velocity at the bottom of the waterfall, which is the horizontal component you have already calculated (4.42 m/s).

You can also use the Pythagorean theorem to calculate the magnitude of the initial velocity, which will be equal to the minimum speed required for the salmon to jump the waterfall. So, using the horizontal and vertical components you have calculated, the minimum speed required for the salmon to jump the waterfall is:

V = sqrt(Vox^2 + Voy^2)
V = sqrt((4.42 m/s)^2 + (4.316 m/s)^2)
V = 6.22 m/s

Therefore, the minimum speed required for the salmon to jump the waterfall is 6.22 m/s.