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?
Vv = vertical velocity, V1 = initial velocity, Dv = vertical distance.
Δt = 2V1Sinθ/(acceleration)
ΔDv = VvΔt + 1/2(acceleration)(Δt^2)
Vv = V1Sinθ
The Attempt at a Solution
First I find the time, since I am given the initial velocity and the launch angle. I get a value of 0.5 seconds. Then I find the vertical velocity by using the equation above, and get about 2.24 m/s. Finally, I use all the information I calculated to find the vertical distance traveled by spring toy, and yield a negative value. What am I doing wrong? I am completely stumped as I haven't been taught any new equations to find the vertical distance. Thanks for all the help!