# Spring toy(Kinematics) check my solution(s)

1. Feb 21, 2012

### supernova1203

1. The problem statement, all variables and given/known data
A spring toy is launched from the ground at 2.3 m/s at an angle of 78 degrees to the ground. What is the maximum height reached by the spring toy?

2. Relevant equations
At first i thought it was asking us to find displacement, i realized later total displacement would be 0, since it eventually comes back down, so i thought, hey theyre just asking us to find max vertical distance right?

so instead of using the following formula to find displacement(I still use it to find time)
Δdv=VvΔt+1/2aΔt^2

I decided to go with d=vt

and i used Vv=v1sin∅

3. The attempt at a solution

Vv=v1sin∅

=2.3sin78

Vv=2.24m/s

then i use this into displacement formula and solve for time.

Δdv=VvΔt+1/2aΔt^2

=2.2Δt+1/2aΔt^2

=2.2Δt+1/2(-9.8)Δt^2

=2.2Δt+(-4.9)Δt^2 (here we factor out Δt)

and we are left with

=2.2+(-4.9)Δt

4.9Δt=2.2

Δt=2.2/4.9

Δt=0.4s

now that we have both time and vertical velocity we can find max distance.

d=vt

=(2.24)(0.4)

d=0.896m

does it look right?

2. Feb 21, 2012

### Staff: Mentor

No, it does not look right. First, when solving for the time you have assumed that the vertical displacement is zero. It is not zero at the top of its trajectory. In fact you do not yet know what the vertical displacement is (it's what you're requested to find). Second, by using the formula d = vt to find displacement, you are ignoring the fact that the velocity changes as the toy travels upwards; it's affected by the acceleration due to gravity, so d = vt doesn't apply.

There are better approaches to the problem involving other kinematic formulas. One such involves the known vertical velocities at launch at at the highest point (apex) of the trajectory, and involve the acceleration and distance... Can you think of the appropriate formula?

3. Feb 21, 2012

### supernova1203

hm..my tutor said that since toy eventually comes down, displacement must be zero, so its asking us to find distance...i guess she was wrong, I suppose iv encountered problems like this before, where we've been asked to find vertical displacement, midway the path of a projectile, in which case we would take time and divide it by 2, and then find vertical displacement, does that apply here?

or perhaps i should just go with earlier approach where i used displacement vertical formula, instead of going with what tutor told me, that we arent looking for vertical displacement.

Last edited: Feb 21, 2012
4. Feb 21, 2012

### Staff: Mentor

You could find the total time from launch to landing and divide by two to find the time at apex. Then apply the formula for vertical motion to find the displacement using that time. Seems like more work than is required if you happen to know either the velocity2 formula or know how to apply conservation of energy.

5. Feb 21, 2012

### supernova1203

hm..i dont mind more work, thanks! :) Atleast i know now what i gotta do