A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bold falls off the?

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A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0s later. What was the rockets acceleration.

The answer i got was 14.715m/s^2 which is wrong.

I thought that the bolts initial velocty should equal the rockets final velocity (when the bolt falls off)

So for the bolt:

V0=(-9.81)(6) – Vf=58.86m/s – 0m/s

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For the rocket:

Vf=V0 + a(4s)

= 0 + 4a

Subbing one into the other:

58.86m/s=4a

a=14.715m/s^2

Any help would be nice.

Ty

1 Answer

  • The initial speed of the bolt is not 58.86 m/s.

    Let a be the acceleration of the rocket.

    During the 4 sec lift off, the rocket has reached a heigth of

    h = (1/2)*a*t^2

    with t=4,

    h = (1/2)*a^16

    h = 8*a

    Its velocity at 4 sec is

    v = t*a

    v = 4*a

    The initial velocity of the bolt is thus 4*a.

    During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,

    h = (1/2)*g*t^2 + V0*t

    Subsituting h0=8*a, t=6 and V0=-4*a into it,

    8*a = (1/2)*g*36 – 4*a*6

    Solving fot a,

    a = 5.52 m/s^2


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