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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 950 m above the earth's surface. The rocket's engines give the rocket an upward acceleration so it moves with acceleration of 16.0 m/s2 during the time t that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance.

Respuesta :

solution:

acceleration of the rocket , a = 16 m/s^2

total height , H = 940 m

let the time taken for rocket to stop after acceleration is t

as 16 \times T = 9.8 \times t

t = 1.63 \times T

Now, for height = 950 m

h = 0.5 \times 16 \times T^2 + 0.5 \times 9.8 \times t^2

950 = 0.5 \times 16 \times T^2 + 0.5 \times 9.8 \times(1.63 \times T)^2

solving for T

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