Calculating Max Height & Travel of Spring

In summary, the problem involved a spring being launched from the ground at an angle of 78° with a velocity of 2.3 m/s. Using the equations for vertical and horizontal components, the vertical velocity was found to be 33.3 m/s and the horizontal velocity was found to be 7.1 m/s. Time was then calculated using the equation for vertical displacement, and it was determined that the maximum height would be achieved at half the total time the spring was in the air. The maximum height was found to be 56.58 m and the total distance traveled by the spring was 48.3 m.
  • #1
STEMucator
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Homework Statement



A spring is launched from the ground at 2.3 m/s t an angle of 78° to the ground. What is the max height of the spring? How far did the spring travel?

Homework Equations



##\vec{v}_R = 34 m/s##
##\theta = 78°##

The Attempt at a Solution



First things first, get my components :

##\vec{v_v} = 34sin(78°) = 33.3 m/s [Up]## [ Vertical velocity ].
##\vec{v_H} = 34cos(78°) = 7.1 m/s [Forward]##

Hmm looks like I need time. So let's use ##\vec{Δd}_V = \vec{v_v}Δt - (4.9)(Δt)^2##. We know that the overall vertical displacement will be 0 because the spring returns to rest due to gravity. So simplifying :

##0 = (33.3 - 4.9Δt)Δt##

So either ##Δt = 0## which we know we can throw out, or ##Δt = \frac{-33.3}{-4.9} = 6.8 s##.

Now we know how long the spring was in the air. So because of acceleration due to gravity, we know that the max height of the spring will be achieved at ##Δt_S = Δt/2 = 3.4 s##.

To find the actual maximum height we calculate :

##\vec{Δd}_V = \vec{v}_v Δt_S - (4.9)(Δt_S)^2 = (33.3)(3.4) - (4.9)(3.4)^2 = 56.58 m/s [Up]##

To find out how far the spring traveled in total, another small calculation :

##\vec{Δd}_H = \vec{v}_H Δt = (7.1)(6.8) = 48.3 m [Forward]##.

I think that should be it. Loving these word problems.

EDIT : Just noticed this should be in intro phys, I must have clicked adv by accident. If a mod could move it there it would be appreciated.
 
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  • #2
Zondrina said:

Homework Statement



A spring is launched from the ground at 2.3 m/s t an angle of 78° to the ground. What is the max height of the spring? How far did the spring travel?

Homework Equations



##\vec{v}_R = 34 m/s##
##\theta = 78°##

The Attempt at a Solution



First things first, get my components :

##\vec{v_v} = 34sin(78°) = 33.3 m/s [Up]## [ Vertical velocity ].
##\vec{v_H} = 34cos(78°) = 7.1 m/s [Forward]##

Hmm looks like I need time. So let's use ##\vec{Δd}_V = \vec{v_v}Δt - (4.9)(Δt)^2##. We know that the overall vertical displacement will be 0 because the spring returns to rest due to gravity. So simplifying :

##0 = (33.3 - 4.9Δt)Δt##

So either ##Δt = 0## which we know we can throw out, or ##Δt = \frac{-33.3}{-4.9} = 6.8 s##.

Now we know how long the spring was in the air. So because of acceleration due to gravity, we know that the max height of the spring will be achieved at ##Δt_S = Δt/2 = 3.4 s##.

To find the actual maximum height we calculate :

##\vec{Δd}_V = \vec{v}_v Δt_S - (4.9)(Δt_S)^2 = (33.3)(3.4) - (4.9)(3.4)^2 = 56.58 m/s [Up]##

To find out how far the spring traveled in total, another small calculation :

##\vec{Δd}_H = \vec{v}_H Δt = (7.1)(6.8) = 48.3 m [Forward]##.

I think that should be it. Loving these word problems.

EDIT : Just noticed this should be in intro phys, I must have clicked adv by accident. If a mod could move it there it would be appreciated.

I'm not sure why your statement says velocity 2.3m/s and you finally used 34m/s. But other than that it looks pretty ok.
 
  • #3
Dick said:
I'm not sure why your statement says velocity 2.3m/s and you finally used 34m/s. But other than that it looks pretty ok.

Oh wow haha. Been awake since 5am, taking its toll i guess, but these word problems are too fun to solve.

Quantifying the physical.

Okay though, as long as the process is fine, that's all I really care for.
 

Related to Calculating Max Height & Travel of Spring

1. How do you calculate the maximum height of a spring?

The maximum height of a spring can be calculated using the formula H = (kx2)/2m, where H is the maximum height, k is the spring constant, x is the displacement of the spring, and m is the mass attached to the spring.

2. What is the spring constant?

The spring constant, also known as the force constant, is a measure of the stiffness of a spring. It is denoted by the letter k and is typically measured in units of newtons per meter (N/m).

3. How do you determine the travel distance of a spring?

The travel distance of a spring can be determined by calculating the displacement of the spring from its equilibrium position. This can be done by measuring the change in length of the spring or by using the formula x = F/k, where x is the displacement, F is the applied force, and k is the spring constant.

4. Can the maximum height and travel distance of a spring be affected by external factors?

Yes, the maximum height and travel distance of a spring can be affected by external factors such as temperature, humidity, and the material of the spring. These factors can alter the spring constant and thus impact the calculations for maximum height and travel distance.

5. What is the significance of calculating the maximum height and travel distance of a spring?

Calculating the maximum height and travel distance of a spring is important in understanding the behavior and characteristics of the spring. It can also be useful in engineering and design applications where the spring is used to store or release energy, such as in a spring-loaded mechanism.

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