TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Let acceleration of rocket, a1=1ms−2,
1 st piece of rocket breaks off after time, t1=20s and s1 be the distance covered by rocket in time t1.
As we know that,
s1=u1t1+‌

1

2

a1t12
where, u1 is the initial speed of rocket while going up.
∴‌‌s1=0×20+‌

1

2

×1×(20)2
s1=200m
and v1=u1+a1t1
where, v1 is speed of rocket after time 20 s.
∴‌‌v1=0+1×20=20ms−1
Now, after 20s the piece of rocket breaks off, the time taken by the piece to reach on ground be t.
Speed of rocket after 20s(v1)= speed of broken part of rocket (u2)
i.e. u2=v1∴‌‌s2=−u2t+‌

1

2

a2t2
⇒‌‌s1=−v1t+‌

1

2

gt2‌‌[∴s1=s2=200m]
⇒‌‌200=−20t+‌

1

2

×10×t2
⇒‌‌40=−4t+t2
⇒‌‌t2−4t−40=0
On solving quadratic equation, we get
t=8.5s or t=−4.63s
Since, time cannot be negative,
∴‌‌t=8.5s