REMARKS You may think that it is more natural to break this problem into three p

Question

REMARKS You may think that it is more natural to break this problem into three phases, with the second phase ending at the maximum height and the third phase a free fall from maximum height to the ground. Although this approach gives the correct answer, it's an unnecessary complication. Two phases are sufficient, one for each different acceleration.

QUESTION If, instead, some fuel remains, at what height should the engines be fired again to brake the rocket's fall and allow a perfectly soft landing? (Assume the same acceleration as in the initial descent.)
m

PRACTICE IT

Use the worked example above to help you solve this problem. A rocket moves straight upward, starting from rest with an acceleration of +28.7 m/s2. It runs out of fuel at the end of 4.66 s and continues to coast upward, reaching a maximum height before falling back to Earth.
(a) Find the rocket's velocity and position at the end of 4.66 s.

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

Explanation / Answer

PRACTICE IT:

(A) vf = vi + a t

vb = 0 + (28.7 x 4.66) = 133.7 m/s

yf - yi = vi t + a t^2 / 2

yb - 0 = (0 x 4.66) + (28.7 x 4.66^2 / 2)

yb = 311.6 m

(B) after 4.66 s, acceleration will be due to gravity.

thatis a = - 9.8 m/s^2

it will untill 133.7 m/s becomes zero.

applying vf^2 - vi^2 = 2 a (yf - yi)

0^2 - (133.7^2) = 2(-9.8)(H - 311.6)

H = 1223.6 m

(C) after that it will fall 1223.6 under the gravity.

v^2 - 0^2 = 2 (-9.8)( 0 - 1223.6)

v = - 155 m/s (velocity will downward )

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EXERCISE :

Applying vf^2 - vi^2 = 2 a (yf - yi)

v^2 - 0^2 = 2 (-9.8) (90.4 - 263)

v = -58.2 m/s (downward )

after that,

0^2 - (-58.2)^2 = 2(a)(0 - 90.4)

a = 18.7 m/s^2 ........ANs

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