(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Spring toy(Kinematics) check my solution(s)

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