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Deffense [45]
3 years ago
6

During a test a rocket travels upward at 90 m/s , and when it is 50 m from the ground its engine fails. Determine the maximum he

ight sb reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m / s 2 due to gravity. Neglect the effect of air resistance.

Physics
2 answers:
Maslowich3 years ago
8 0

Answer:

Maximum height sb = 462.84m

Speed before it hits the ground v = 95.29 m/s

Explanation:

Given;

Initial height before engine fails h1 = 50 m

Rocket speed before engine fail u =90m/s

Acceleration due to gravity g = 9.81m/s^2

Maximum height from point of engine fail h2;

h2 = u^2/2g = 90^2/(2×9.81)

h2 = 412.84 m

Maximum height sb = h1 + h2 = 50m + 412.84m

sb = 462.84m

The speed before touching the ground v;

v = √2gsb = √(2×9.81×462.84)

v = 95.29 m/s

katrin [286]3 years ago
6 0

Answer:

Maximum height above ground = 462.8m

Speed before hitting ground = 95.3m/s

Explanation:

Detailed explanation and calculation is shown in the image below

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