# Maximum height of a spring toy

## 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.

SteamKing
Staff Emeritus
Homework Helper

## 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?

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?

SteamKing
Staff Emeritus
Homework Helper
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.

Much better.

Thanks for the help :)