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?