A rocket lifts off from the ground and accele | vedclass

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(D) $1$. Calculate the height of the rocket at $t = 20 \ s$: $h = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 1 	imes (20)^2 = 200 \ m$.$2$. Calculate the

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(D) $1$. Calculate the height of the rocket at $t = 20 \ s$: $h = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 1 	imes (20)^2 = 200 \ m$.$2$. Calculate the velocity of the rocket at $t = 20 \ s$: $v = a t = 1 	imes 20 = 20 \ m \ s^{-1}$.$3$. The piece breaks off and acts as a projectile with initial velocity $u = 20 \ m \ s^{-1}$ upwards from a height $h = 200 \ m$.$4$. Using the equation of motion $s = ut + \frac{1}{2} a t^2$ for the piece, where $s = -200 \ m$ (downward displacement), $u = 20 \ m \ s^{-1}$, and $a = -g = -10 \ m \ s^{-2}$:$-200 = 20 t - \frac{1}{2} 	imes 10 	imes t^2$$-200 = 20 t - 5 t^2$$5 t^2 - 20 t - 200 = 0$$t^2 - 4 t - 40 = 0$.$5$. Solving for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{4 \pm \sqrt{16 - 4(1)(-40)}}{2} = \frac{4 \pm \sqrt{176}}{2} = 2 \pm \sqrt{44} \approx 2 \pm 6.63$.$6$. Since time must be positive, $t \approx 8.63 \ s$.

$A$ rocket lifts off from the ground and accelerates upwards at $1 \ m \ s^{-2}$. $20 \ s$ after liftoff, a piece breaks off from the bottom of the rocket. After breaking off, how much time does it take approximately to reach the ground (in $s$)? (Take $g = 10 \ m \ s^{-2}$)

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