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17 November, 20:17

a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of the rocket and hits the ground 6.00 seconds later. what was the rockets acceleration?

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  1. 17 November, 22:52
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    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 height 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

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

    8*a = (1/2) * g*36 - 4*a*6

    Solving for a

    a = 5.52 m/s^2
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