Sarthaks APP
+2 votes
in Physics by (7.5k points)
A rocket is fired vertically from the ground. It moves upwards with a constant acceleration of 10ms^-2 for 30 seconds after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height? (g=10 m/s^=-2)

1 Answer

+3 votes
by (13.5k points)
selected by
Best answer


The distance travelled by the rocket during consumption of fuel ( 30 sec.) in which resultant acceleration is vertically upwards i.e 10 m/s2 will be

h = ut + 1/2gt2

h = 0*30 + 1/2*10*30= 4500m --------------------(1)

Velocity will be

v = u +at

v = 0 + 10*30 = 300m/s ---------------(2)

As after the fuel is consumed the initial velocity from eqn. (2) is 300 m/s and gravity opposes the motion of rocket, so from 1 st equation of motion time taken by it to reach the  maximum height (for which v = 0)

  v = u – gt

  0 = 300 – 10t

  t = 30 sec

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.