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Home / Doubts and Solutions/JEE/Physics

how to do this numerical??

 

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Asked by pukai7229 23rd June 2018, 10:09 AM
Answered by Expert
Answer:
Initial height h1 travelled with fuel is calculated using the formula " S = u×t+(1/2)a×t2 " ,
with initial velocity u =0,  a = 20  m/s2  and time t = 60 s.
 
h1 = (1/2)×20×60×60 = 36000 m = 36 km ...........................(1)
 
speed v after 1 min is calculated using the formula " v = u+a×t " ,
with initial speed u =0, acceleration a = 20 m/s2 and time t = 60s.
 
v = 20×60 = 1200 m/s.
 
height h2 reached by rocket after completion of fuel is calculated using the formula " v2 = u2 - 2×g×s "
with  final speed v =0, inital speed u =1200 m/s and s = h2 . let us assume g = 10 m/s2
 
h2 = (1200×1200) /  ( 2 × 10 ) = 72000 = 72 km. ...........................(2)
 
maximum height from starting point = 72+36 = 108 km
 
time t2 to reach maximum height is calculated using the formula, " v = u - gt ", with final speed v=0.
 
t2 = 1200/10 = 120 s ........................(3)
 
Time t3 to reach ground is calculated by formula, " S = u×t + (1/2)g×t2 ", with initial speed u = 0
and  S = maximum height from starting point = 72+36 = 108 km
 
108×1000 = (1/2)×10×t32 .......................(4)
 
solving eqn. (4) for t3,  we get t3 = 147 s
 
total time = initial time 60 s+ t2 + t3 = 60+120+147 = 327 s
 
 
Answered by Expert 23rd June 2018, 2:24 PM
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