Question

1) A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket at 36 m above ground is closest to

Answer 1

Using conservation of energy

\[K_{1}+U_{1}=K_{2}+U_{2}\]
we can let K1 be the kinetic energy at the ground, U1 be the potential energy at the ground, K2 be the kinetic energy at the burnout point, and U2 be the potential energy at the burnout point.

Substituting the appropriate formulas:
\[\frac{ 1 }{ 2 }mv_{1}^{2}+mgh_{1}=\frac{ 1 }{ 2 }mv_{2}^{2}+mgh_{2}\]
at the ground, h1 = 0. v1 is unknown. v2 = velocity at burnout = 30 m/s, h2 = 72 m. the mass of the rocket is constant so m can cancel out on each side. acceleration due to gravity = 9.81 m/s^2.
plugging in these quantities:
\[\frac{ 1 }{ 2 }v_{1}^{2}=\frac{ 1 }{ 2 }(30)^{2}+(9.81)(72)\]
you can solve for v1. once you have v1 you can set up the conservation equation again. the left side can once again be the rocket on the ground, but on the right side you can set h2 = 36 m and solve for v2.