A rocket lifts off from ground and accelerate upwards at $1 ms^{- 2} 20 s$ 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 $g = 10 ms^{- 2}$ )

Question

A rocket lifts off from ground and accelerate upwards at

1 ms^{- 2} 20 s

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 = 10 ms^{- 2}

)

Accepted Answer

Correct answer: D. (d) Distance covered by rocket in t=20 s, ∵s= ut+ 12at^2 ⇒s=0 20+ 12 1 (20 )^2=200m Now using v=u+ at where, v is speed of rocket after time 20 s. v=0+1 20=20 ms-1 Now, after 20 s the piece of rocket breaks off, the time taken by the piece to reach on ground be t'. Speed of rocket after 20 s (v )= speed of broken part of rocket (u^ ' ) ) i.e. u^ '=v u^ '= va 'g ∵s^ '=-u^ 't+ 12a^ ' (t^ ' )^2( ∵s^ '=s=200 m)a=1 ms-2u=0 ⇒200=-20t+ 12 10 t^2 ∵40=-4t+t^2 ⇒t^2-4t-40=0 Solving we get t=8.5 s ort=-4.63 s Since, time cannot be negative, so t=8.5 s

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

Solution

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