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

AI Thread Summary
The maximum height of a spring toy launched at an angle of 78° with an initial velocity of 2.3 m/s was initially calculated incorrectly without considering the vertical component of the velocity. After recalculating using the vertical component, which is approximately 2.25 m/s, the correct maximum height reached was determined to be 0.26 m. The calculations involved using the equation for vertical motion, factoring in acceleration due to gravity. The discussion highlights the importance of using the angle to find the vertical velocity component for accurate results. Overall, the final answer of 0.26 m is confirmed as correct.
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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