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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)
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*t²h = 0* 60 + 1/2 *10 *60²h = 1800 m Now the velocity acquired will be-V = u + a*tV = 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
The distance travelled by the rocket in 1 minute (60 s) with a=10m/s2a=10m/s2 upward ish1=12h1=12×10×602×10×602=18000m=18000m=18km=18kmVelocity acquired =>v=10×60v=10×60=600m/s=600m/safter 1 minute the rocket moves upward with velocity 600m/s600m/s and down ward.acceleration due to gravity till its velocity becomes zero0=(600)2−2gh20=(600)2−2gh2=(600)2−2×10×h2=(600)2−2×10×h2=>h2=18000m=>h2=18000m=>18km=>18kmTotal distance 18+18=36km
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