# Maximum height of a spring toy launched at an angle

• CAT 2
In summary: You could derive an expression for the time using the following equation: Δt = 2(V1)Sinθ/(acceleration). In summary, the spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. At the maximum height reached by the spring toy, it has traveled for 0.46s.
CAT 2

## 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

Δt = 2(V1)Sinθ/(acceleration)
ΔD = V1Δt + 1/2(acceleration)(Δt^2)

## The Attempt at a Solution

First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)
Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)
Δt = 0.46s

Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s)0.46s + 1/2(9.8 m/s)(0.46^2)
ΔD = 1.058 + 1.037
ΔD = 2.095m

Is this the right answer? Is this the right equation or should I be using ΔDv = 1/2(acceleration)Δt^2 ?Somehow this doesn't seem right because I don't know when the toy is at the highest point. If I knew the time of its highest point then I could find the vertical distance using that time, that is how I have been taught.

One more thing, should I be saying [up] is positive and using - 9.8 m/s[up] in my above equations?
e.g. ΔD = (2.3m/s[up])0.46s + 1/2(-9.8 m/s[up])(0.46^2)

Thanks for any help

Hi Cat 2,

Welcome to Physics Forums.

You might notice that I've changed the title of your thread to make it descriptive of the problem being posed. You want to use highly descriptive titles in order to grab the attention of the Homework Helpers with the right skill set

In your attempt, what time is being calculated by the formula that you've employed? Are you sure that the maximum height occurs at that time?

Thanks gneill,

OK, I'm not sure that us where the maximum height occurs. I don't have a formula for finding maximum height nor have I had a question that asked this before. I said above...
CAT 2 said:
Is this the right answer? Is this the right equation or should I be using ΔDv = 1/2(acceleration)Δt^2 ?Somehow this doesn't seem right because I don't know when the toy is at the highest point. If I knew the time of its highest point then I could find the vertical distance using that time, that is how I have been taught.

That is what I have been taught to do.

The formula I used, found the time of the flight, 0.46s. Should I be saying that the highest the toy goes would be half of the time of the flight, 0.23s? (Half of a parabola is its maximum point and vertex.)

CAT 2 said:
The formula I used, found the time of the flight, 0.46s. Should I be saying that the highest the toy goes would be half of the time of the flight, 0.23s? (Half of a parabola is its maximum point and vertex.)
Yes!

Alternatively, you could derive an expression for the time of maximum height by writing the equation of motion for the y-component and using a bit of calculus to find when the y-value is maximized.

OK, so would my answer look like this then?

First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)
Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)
Δt = 0.46s

0.46s/0.5 = 0.23s - this is half of the total time of my parabola, therefore it will be the highest point.

Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s)0.23s + 1/2(9.8 m/s)(0.23s^2)
ΔD = 0.529 + 0.259
ΔD = 0.788 meters

CAT 2 said:
OK, so would my answer look like this then?

First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)
Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)
Δt = 0.46s

0.46s/0.5 = 0.23s - this is half of the total time of my parabola, therefore it will be the highest point.

Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s)0.23s + 1/2(9.8 m/s)(0.23s^2)
ΔD = 0.529 + 0.259
ΔD = 0.788 meters

Not quite. Two things here,

Thinking generally about the physics here, do you expect that the acceleration due to gravity will have a positive contribution to the max upward distance or negative one? Is that in agreement with your calculations? What are the directions here?

What is the 2.3sin(78) in the equation for t? Do you understand why that is in the equation for t? Should it appear in the equation for D?

Thanks for helping RedDelicious!
My acceleration should've been -9.8 m/s^2 [up], and the speed, 2.3m/s [up].

Also (2.3m/s[up])0.23s should have been (2.3m/s[up])(sine78)0.23s

ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s[up])(sine78)0.23s + 1/2(-9.8 m/s^2[up])(0.23s^2)
ΔD = 0.517 - 0.259
ΔD = 0.258 meters

How's that?

Did I do that correctly now?

Good attem
CAT 2 said:
Thanks for helping RedDelicious!
My acceleration should've been -9.8 m/s^2 [up], and the speed, 2.3m/s [up].

Also (2.3m/s[up])0.23s should have been (2.3m/s[up])(sine78)0.23s

ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s[up])(sine78)0.23s + 1/2(-9.8 m/s^2[up])(0.23s^2)
ΔD = 0.517 - 0.259
ΔD = 0.258 meters

How's that?
Good!

A couple of technical points about the presentation of your equations. If you start by defining your coordinate system, you don't have to label each value with [up] (or [down]). The appropriate direction is implied by the sign of the value once the coordinate system is defined. Thus:

Assuming up is the direction of the positive y-axis:

ΔD = (2.3m/s)(sine78)0.23s + 1/2(-9.8 m/s^2)(0.23s)^2

Note also that I moved the final parenthesis so that the entire time quantity, including its units, are enclosed for the squaring operation. Otherwise one might misinterpret things, supposing that only the units were to be squared.

Alright, I think I got it all! Thanks a lot for all your help gneill!

CAT 2 said:
Δt = 2(V1)Sinθ/(acceleration)
I recommend you delete this equation from your notes. The trouble with long lists of memorised equations is that you forget which apply when. Better to keep a short list of the most general equations.
The one you quote is for the specific case of time to return to the same height.

There are five "SUVAT" equations that cover all the constant acceleration problems you will encounter. The acronym represents the five variables, s=displacement, u= initial velocity, v=final velocity, a=acceleration, t=time. Each equation omits one variable and relates the other four; identify which three you know (or connect to other facts), which one is to be determined, and select the equation involving those four.

See e.g. https://en.m.wikipedia.org/wiki/Equations_of_motion

In the present case, you initial vertical velocity, vertical acceleration, and final vertical velocity (0 at highest point) and you wish to find the time.

Wait, as I am fixing this problem and adding corrections (-9.8 m/s [up] and 2.3 m/s [up]), I find that if I apply this to the second equation I get a positive answer...
ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s[up])(sine78)0.23s + 1/2(-9.8 m/s^2[up])(0.23s^2)
ΔD = 0.517 - 0.259
ΔD = 0.258 meters

But if I apply this to the first equation and then the second equation, I get a negative answer...
Δt = 2(V1)Sinθ/(acceleration)
Δt = 2(2.3 m/s) Sin78 / (-9.8 m/s)
Δt = -0.46s

ΔD = V1Δt + 1/2(acceleration)(Δt^2)
ΔD = (2.3m/s)(sine78)-0.23s + 1/2(-9.8 m/s^2)(-0.23s)^2
ΔD = - 0.517 - 0.259
ΔD = - 0.776 meters

Do I a) did something wrong. (I don't think so.)
b) ignore (remove) the negative sign and this is my answer. (Maybe? But that doesn't make sense because the - has a purpose. )
c) not use (-9.8 m/s [up] in the first equation. ( Maybe, because I am just finding time and not distance so it doesn't matter?)

What do I do to not get a negative answer? use c) ?

You need to look at the derivation of your formula:
CAT 2 said:
Δt = 2(V1)Sinθ/(acceleration)
in order to make sure that you understand where and how to apply it. Applying formulas without really understanding where they come from is not recommended. It can lead to unexpected errors.

Can you derive that equation from the general motion equation ##s = v\;t + \frac{1}{2} a t^2##? I think you'll spot the problem you had if you do.

No, unfortunately I completely don't get what deriving the equation from the general motion equation has to do with the problem. Could you help me out a little more? Does it have to do with what your finding e.g. if your finding displacement then you need directions, but if your finding time, not?

CAT 2 said:
No, unfortunately I completely don't get what deriving the equation from the general motion equation has to do with the problem. Could you help me out a little more? Does it have to do with what your finding e.g. if your finding displacement then you need directions, but if your finding time, not?
You are blindly applying a formula with unwarranted assumptions about the variables in it. Hence, a sign error has occurred giving you an incorrect (negative) value for the time.

Deriving the equation yourself would give you insight into the issue, and further, it would be one less equation to memorize for your tests and exams: you could easily derive it in a matter of moments from the basic equation of motion, and be certain of what the variables mean and what signs are appropriate.

Ok, I got this, in the first equation the - in 9.8 m/s is already in, don't need o put in another one!

Thanks for all the help

## 1. How does the angle at which a spring toy is launched affect its maximum height?

The angle at which a spring toy is launched determines the initial velocity and direction of the toy. This, in turn, affects the toy's maximum height. A higher launch angle will result in a higher initial velocity and, therefore, a higher maximum height. However, if the launch angle is too high, the toy may not have enough horizontal velocity to reach its maximum height and will instead fall shorter.

## 2. Is there a mathematical formula to calculate the maximum height of a spring toy launched at an angle?

Yes, there is a formula that can be used to calculate the maximum height of a spring toy launched at an angle. The formula is h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s2).

## 3. What factors besides the launch angle can affect the maximum height of a spring toy?

Besides the launch angle, the maximum height of a spring toy can also be affected by the initial velocity, the mass of the toy, and the air resistance. A higher initial velocity and a lighter toy will result in a higher maximum height. Air resistance can also play a role, as it can slow down the toy and reduce its maximum height.

## 4. How does the surface on which a spring toy is launched affect its maximum height?

The surface on which a spring toy is launched can affect its maximum height in two ways. Firstly, a rough or uneven surface can cause the toy to lose some of its initial velocity, resulting in a lower maximum height. Secondly, a surface with a different angle or slope can change the launch angle of the toy, which will also affect its maximum height.

## 5. Can the maximum height of a spring toy launched at an angle be greater than the height at which it is released?

Yes, it is possible for the maximum height of a spring toy launched at an angle to be greater than the height at which it is released. This is because the initial velocity and launch angle can give the toy enough energy to overcome the effects of gravity and reach a higher maximum height. However, this is only possible if the launch angle and initial velocity are within certain limits and if there is no significant air resistance.

• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
5K
• Introductory Physics Homework Help
Replies
10
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
7K
• Introductory Physics Homework Help
Replies
26
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
2K