Question

A rocket when fired rises uo with an acceleration of 20 m/s^2 for the first 1.0 min.after this its fuel is finished and it goes on rising up just like a free body. a)up to what maximum height will it rise? b)how much total time will it take before falling on earth?

Answer 1

Velocity for 20m/s² accelration
V1= u + at
= 0 + 20 x 1 60 
=1200 m/s

Height = h = ut + 1/2 + 20 + 60²
= 36000m

Velocity for 10 m/s²
v2 = v1 + gt'
0 = 1200 - 10t
t' = 120 s

Height for 10 m/s²

h'= v1t' + 1/2gt'²
= 1200 x 120 -1/2 x 10 x 120²
= 72000m

total height = h+h'
= 36000 + 72000
= 108000m

Time taken 
h=1/2 gt²
t = √2hg
= √2 x 108000 x 10
= 600√6 sec or =60 (3+√6)

Answer 2

velocity at the end of 1.0 min = v = at = 20 * 60 = 1,200 m/s
Height reached by t = 1.0 min => h1 = 1/2 a t² 
    h1 = 1/2 * 20 * 60² = 36, 000 m 

After 1.0 minutes, time to reach the highest point : t
     t = v / g = 1200/10 = 120 sec
height reached during 120 sec = h2 = v² / 2g = 1200²/ 2*10 = 72,000 m

Maximum height reached = h1 + h2 = 108 km

time to fall on Earth from top point : t
          t = √[2 (h1+h2)/g = √(2*108000/10) = 60 √6 sec

Total time of flight = 60 + 120 + 60 √6 sec = 60 (3 + √6) sec