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/sFor the rocket:Vf=V0 + a(4s) = 0 + 4aSubbing one into the other:58.86m/s=4aa=14.715m/s^2Any help would be nice.Ty

Hermina Ferry · Accepted 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^2with t=4,h = (1/2)*a^16h = 8*aIts velocity at 4 sec isv = t*av = 4*aThe 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*tSubsituting h0=8*a, t=6 and V0=-4*a into it,8*a = (1/2)*g*36 – 4*a*6Solving fot a,a = 5.52 m/s^2

Jakob Littel

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

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