What is the maximum height of a spring toy launched at an angle?

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Homework Help Overview

The discussion revolves around determining the maximum height of a spring toy launched from the ground at a speed of 2.3 m/s and an angle of 78° to the ground. Participants are exploring the physics of projectile motion, specifically focusing on the vertical component of the initial velocity and the equations of motion involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants initially calculate the maximum height using the total velocity without considering the angle, leading to questions about the correctness of their approach. Subsequent attempts involve calculating the vertical component of the initial velocity using trigonometric functions, prompting further discussion on the accuracy of their results.

Discussion Status

The discussion has evolved with participants revisiting their calculations after recognizing the importance of the vertical component of the initial velocity. There is acknowledgment of improved calculations, but no explicit consensus on the final answer has been reached.

Contextual Notes

Participants express uncertainty about their calculations and the necessity of incorporating the launch angle into their analysis. The repeated emphasis on the vertical component suggests a focus on understanding the underlying physics rather than simply arriving at a numerical answer.

Evangeline101
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Homework Statement


A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

Homework Equations



Δ[/B]dv = v1Δt + 1/2aΔt2

The Attempt at a Solution


[/B]
Calculate time:

Δt = 2.3 m/s / 9.8m/s2

= 0.23 s

Now calculate vertical motion:
Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

Δdv = (2.3 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

Δdv = 0.27 m [up]

The maximum height reached by the spring toy is 0.27 m [up].

Is this correct? I have a feeling I did it wrong because I did not use the angle they provided and only used the velocity.
 
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Evangeline101 said:

Homework Statement


A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

Homework Equations



Δ[/B]dv = v1Δt + 1/2aΔt2

The Attempt at a Solution


[/B]
Calculate time:

Δt = 2.3 m/s / 9.8m/s2

= 0.23 s

Now calculate vertical motion:
Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

Δdv = (2.3 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

Δdv = 0.27 m [up]

The maximum height reached by the spring toy is 0.27 m [up].

Is this correct? I have a feeling I did it wrong because I did not use the angle they provided and only used the velocity.

No, it's not correct. Why didn't you calculate the vertical component of the initial velocity?
 
SteamKing said:
No, it's not correct. Why didn't you calculate the vertical component of the initial velocity?

Okay, so I did it again with the vertical component of initial velocity.

Calculate vertical component of initial velocity:

V1v = v1sinθ

V1v = (2.3 m/s) sin 78°

V1v = 2.25 m/s [up]

Calculate time:

Δt = 2.3 m/s / 9.8m/s2

= 0.23 s

Calculate vertical motion:
Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

Δdv = (2.25 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

Δdv = 0.26 m [up]

The maximum height reached by the spring toy is 0.26 m [up].

Is this better?
 
Evangeline101 said:
Okay, so I did it again with the vertical component of initial velocity.

Calculate vertical component of initial velocity:

V1v = v1sinθ

V1v = (2.3 m/s) sin 78°

V1v = 2.25 m/s [up]

Calculate time:

Δt = 2.3 m/s / 9.8m/s2

= 0.23 s

Calculate vertical motion:
Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive

Δdv = (2.25 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2

Δdv = 0.26 m [up]

The maximum height reached by the spring toy is 0.26 m [up].

Is this better?
Much better.
 
SteamKing said:
Much better.

Thanks for the help :)
 

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