Answer:
the release will be at 3.266 m distance
Explanation:
mass = 1 Kg
spring constant (k) = 800 N/m
initial compression = 0.20 m
θ = 30⁰
[tex]U= \dfrac{1}{2}kx^2\\U= \dfrac{1}{2}800\times 0.2^2\\U= 16 J[/tex]
[tex]U=mgh\\h=\dfrac{U}{mg}\\h=\dfrac{16}{1 \times 9.8}\\h = 1.633m[/tex]
[tex]d=\dfrac{h}{sin \theta}\\d=\dfrac{1.633}{sin 30}\\\\d= 3.266m[/tex]
hence the release will be at 3.266 m distance.