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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 upon impact on the ground is closest to ...?

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The toy rocket is launched vertically from ground level, 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 total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2

=> v^2 = 11556/5

=> v = 48.07

The velocity of the rocket when it impacts the ground is 48.07 m/s

oladayoademosu

The speed of the rocket upon impact on the ground is closest to 48.08 m/s

Conservation of energy

From the law of conservation of energy, the initial mechanical energy at 72 m, E equals the final mechanical energy at the ground level, E'.

So, E = E'

mgh + 1/2mv² = mgh' + 1/2mv'² where

  • m = mass of toy rocket,
  • g = acceleration due to gravity = 9.8 m/s²,
  • h = initial height of rocket = 72 m,
  • v = initial speed of rocket = 30 m/s,
  • h = final height of rocket = 0 m(since it is ground level) and
  • v' = final speed of rocket

So, mgh + 1/2mv² = mgh' + 1/2mv'²

gh + 1/2v² = gh' + 1/2v'²

Speed of rocket

Making v' subject of the formula, we have

v' = √[v² + 2g(h - h')]

Substituting the values of the variables into the equation, we have

v' = √[v² + 2g(h - h')]

v' = √[(30 m/s)² + 2 × 9.8 m/s²(72 m - 0 m)]

v' = √[900 m²/s² + 19.6 m/s² × 72 m]

v' = √[900 m²/s² + 1411.2 m²/s²]

v' = √[2311.2 m²/s²]

v = 48.075 m/s

v ≅ 48.08 m/s

So, speed of the rocket upon impact on the ground is closest to 48.08 m/s

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