Answer to Question #231172 in Classical Mechanics for Sridhar

Question

A rocket lifts off from ground and accelerate upwards at 1ms^(2).20 seconds after lift off a piece breaks off from the bottom of rocket.After breaking off how much time it takes approximately to reach the ground? (Take g=10ms^(2) )
Ans 8.5 s

Answer 1

$υ = υ_0+ at$

$y = y_0 + v_0t+\frac{at^2}{2}$

$\text{For a rocket through }t=20s$

$a = 1\frac{m}{s^2}$

$v_0 = 0$

$v_r = 1*20= 20\frac{m}{s}$

$y_0 = 0$

$y_r = \frac{1*20^2}{2}=200m$

$\text{For a piece of rocket}$

$a =- g = -10\frac{m}{s^2}$

$v_0= v_r = 20\frac{m}{s}$

$y_p = 0$

$y_0 = y_r = 200m$

$y = y_0 + v_0t+\frac{at^2}{2}$

$0 = 200 +20t-\frac{10*t^2}{2}$

$t^2- 4t-40 = 0$

$t_1 \approx 8.63 s$

$t_2 \approx -4.63 <0$

$\text{Answer: }8.63s$