EARN MORE REMARKS You may think that it is more natural to break this problem in

Question

EARN MORE

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

(A) v= u + at and yf - yi = v0 t + a t^2 /2  

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

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

h = yb = 311.6 m

(B) maximum height will be when v = 0

vf^2 - vi^2 = 2 a y

0^2 - 133.7^2 = (2 x -9.81 x y)

y = 912 m  

Hmax= y + yb = 1223.6 m

(C) vf^2 - vi^2 = 2 a y

vf^2 - 0^2 = 2(-9.81)( 0 - 1223.6)

vf = - 155 m/s
-----------------------------

first we we have to find speed at 90.4 m height.

vf^2 - vi^2 = 2 a y

v^2 - 0^2 = 2(-9.81)(90.4 - 263)

v = 58.2 m/s

now applying same equation for h - 90.4 m to h = 0

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

a = 18.7 m/s^2

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