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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 78 m and acquired a velocity of 40 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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Answer:

velocity of rocket at the time of impact is 48.28 m/s

Explanation:

given,

rise of the rocket = 78 m

Velocity acquires by the rocket = 40 m/s

final velocity  = 0 m/s

v = u + gt

0 = 40 - 9.81 × t

t = 4.08 s

now, calculation of displacement

[tex]s = u t + \dfrac{1}{2}gt^2[/tex]

[tex]s = 30\times 4.08 - \dfrac{1}{2}\times 9.81\times 4.08^2[/tex]

  = 40.8 m

total height = 40.8 + 78 = 118.8 m

velocity = [tex]\sqrt{2gh} = \sqrt{2\times 9.81\times 118.8 }[/tex]

v = 48.28 m/s

velocity of rocket at the time of impact is 48.28 m/s

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