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Maximum height of a spring toy

  • #1

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.
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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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.
No, it's not correct. Why didn't you calculate the vertical component of the initial velocity?
 
  • #3
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?
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
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.
 
  • #5

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