Jason launches a model rocket with a mass of 1.2 kg from his spring-powered rocket launcher with a spring constant of 800 N/m. He pulls it back .55 m. If Jason aims it straight up,

what will its kinetic energy be immediately after it is released from the spring? J,

What is its maximum velocity? m/s

Through conservation of mechanical energy, we know that the spring potential energy of the launcher will equal the kinetic energyy of the rocket at the peak of its flight. Thus SPE = GPE = 1/2 kx^{2} = 1/2mv^{2} . We know that k = 800, m = 1.2, x = .55.

Thus, we can plug these numbers into the equation and get

GPE = 1/2 (800) (.55)^{ 2} = 121 J.

Thus, the maximum kinetic energy is also 121 J. It’s velocity is given in the equation GPE = KE = 1/2mv^{2}.

We know KE, and m, so we plug them into the equation and get 121 = (1/2)(1.2)v^{2}, v = 14.2 m/s