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Grade: 11

                        

A rocket is fired vertically up from the ground with a resultant vertical acceleration of 10m/sec2.The fuel is is finished in 1min and the rocket continues to move up. After how much time from then will the maximum height be reached?Give calculations.(g=10m/sec2)

3 years ago

Answers : (2)

Arun
24742 Points
							
The distance travelled by the rocket during burning interval (1 min = 60 sec.) in which resultant acceleration is vertically upwards and 10 m/s2 will be
 
h = u*t + 1/2 g*t²
h = 0* 60 + 1/2 *10 *60²
h = 1800 m
 
Now the velocity acquired will be-
V = u + a*t
V = 0+ 10 *60 = 600 m/sec
 
As after burning of fuel the initial velocity from (ii) is 600 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 = 600 – 10t
  t = 60 sec
3 years ago
ankit singh
askIITians Faculty
596 Points
							
The distance travelled by the rocket in 1 minute (60 s) with a=10m/s2a=10m/s2 upward is
h1=12h1=12×10×602×10×602
=18000m=18000m
=18km=18km
Velocity acquired =>
v=10×60v=10×60
=600m/s=600m/s
after 1 minute the rocket moves upward with velocity 600m/s600m/s and down ward.
acceleration due to gravity till its velocity becomes zero
0=(600)2−2gh20=(600)2−2gh2
=(600)2−2×10×h2=(600)2−2×10×h2
=>h2=18000m=>h2=18000m
=>18km=>18km
Total distance 18+18=36km
 
3 months ago
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