AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(a) Resultant vertical acceleration,
a=10ms−2
Height travelled by the rocket in 1 min (60s),
h‌‌=ut+‌

1

2

at2
‌‌=0×60+‌

1

2

×10×(60)2
‌‌=18000m=18km
Velocity of rocket after 1 min (60s),
v=u+at=0+10×60=600ms−1
After 1 min, when fuel of rocket is finished, then
u=600ms−1,a=−g=−9.8m∕s2
Hence, if rocket goes at height h2, then
v2−u2‌‌=2gh2
0−(600)2‌‌=2(−9.8)h2
h2‌‌=‌

360000

2×9.8

‌‌=18367.3m=18.4km
∴ Maximum height travelled by rocket from ground =h1+h2=18+18.4=36.4km