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?
Δ[/B]dv = v1Δt + 1/2aΔt2
The Attempt at a Solution
Δt = 2.3 m/s / 9.8m/s2
= 0.23 s
Now calculate vertical motion:
Δdv = v1Δt + 1/2aΔt2 *for vertical motion, let (up) be positive
Δdv = (2.3 m/s) x (0.23s) + 1/2 (-9.8 m/s2) x (0.23s)2
Δdv = 0.27 m [up]
The maximum height reached by the spring toy is 0.27 m [up].
Is this correct? I have a feeling I did it wrong because I did not use the angle they provided and only used the velocity.