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A rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of 19.0 m/s2. However, after 49.5 s of flight the engine fails. What is the rocket's altitude when the engine fails? Assume g = 9.80 m/s2.

When does it reach its maximum height?

What is the maximum height reached?

What is the velocity of the rocket just before it hits the ground?

Respuesta :

Answer:

total time to reach max height = T +t =95.9 + 49.5 = 145.4 s

max height  = s + h= 23277.4 + 45083.6 = 68361.00m

V = sqrrt(2*9.81*68361) = 1158m/s

Explanation:

initial velocity = u = 0

final velocity = v = ?

a = 19m/s^2

t = 49.5 s

when the engine stops the only acceleration is the gravitational acceleration

so

g = 9.81 m/s^2

height achieved till the engine works = s =u*t + 0.5 *a*t^2

s= 0 +0.5*19*49.5^2

s = 23277.4 m

Vs = velocity at  s = 23277.4 while going up

Vs = u + a*t

Vs = 19 * 49.5 = 940.5 m/s

further height  = h

0^2 - Vs^2 = 2*g*h= 2 * 9.81 * h

940.5^2 = 2*9.81*h

h= 45083.6 m

max height  = s + h= 23277.4 + 45083.6 = 68361 m

T: time to reach max height after engine fails

0-Vs = g * T

940.5/9.81 = T = 95.9s

total time to reach max height = T +t =95.9 + 49.5 = 145.4 s

VELOCITY WHEN IT HITS THE GROUND:

V= ?

U = 0

A = g = 9.81

s = 68361 m

V^2 - U^2 = 2 *9.81 * 68361

V = sqrrt(2*9.81*68361) = 1158m/s

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