Question

Question #754ff

Answer 1

Answer
Given
`u->"Initial velocity of rocket"=0m/s`

`a->"Acceleration of the rocket"=2.10ms^-2`
`t->"Duration of upward acceleration"=30s`

Velocity gained in 30s is

`v=u+axxt=0+2.1*30=63m/s`

1.The engine fails at this moment and rocket undergoes free fall under gravity.If it reaches height h at this point then

`h=uxxt+1/2xxaxxt^2`
`=0*30+0.5*2.1*30^2=945m`

After 30s of its journey it will go up with retardation `g=-9.8ms^-2` till its velocity becomes zero.

2.At this highest point it will have a downward acceleration `g=9.8ms^-2` and vlocity`=63m/s`
If it ascends a height `h'`during last phase of its ascent then
`0^2=63^2-2*9.8*h'`
`h'=63^2/(2*9.8)=202.5m`

If the rocket takes Ts to fall back at launch pad after its engine fails then

Displacement during Ts is `=-h=-945m`
Initial velocity upward `=+63m/s`
And `g=-9.8ms^-2`

So

`-945=63*T-1/2*9.8*T^2`

`=>4.9T^2-63T-945=0`

`=>T=(63+sqrt(63^2+4*4.9*945))/(2*4.9)`
`=>T=21.73s`

So the total time to return back at the launch pad after it was launched will be

3.`"Total time"=T+30=21.73+30=51.73s`

Answer 2

1.
Given Initial velocity of rocket `u=0ms^-1`
Upwards Acceleration of the rocket `a=2.10ms^-2`
Duration of upward acceleration `t=30s`

Velocity `v` at the end of `30s`, assuming upwards direction is positive.
`v=u+at`
`=>v=0+2.10xx30=63ms^-1`

After engine fails the rocket undergoes free fall under gravity with initial velocity calculated above. Let it reach height `h` when engine fails. Using the kinematic equation
`h=ut+1/2at^2`
`=0xx30+1/2xx2.10xx30^2=945m`
2. Thereafter, first it goes up and is retarded under gravity `g=-9.81ms^-2` till its velocity becomes zero. Then there is free fall under gravity.
`:.`At its highest point it has velocity`=0` and acceleration `=-9.81ms^-2`
3. Suppose the rocket takes `t_f` to fall back at launch pad after its engine failed, we have

Displacement during time `t_f`
`="Final position"-"Initial position"=0-945=-945m`
Setting up the kinematic equation we have
`s=ut+1/2at^2`
`=>-945=63xxt_f+1/2xx(-9.81)xxt_f^2`

`=>4.905t_f^2-63t_f-945=0`
Using the quadratic formula
`t_f=(63+-sqrt(63^2+4xx4.905xx945))/(2xx4.905)`
Ignoring the `-ve` root as time can not be negative
`=>t_f=21.7s`, rounded to one decimal place

`:.` the total time to return to launch pad after launch
`=t_f+30=21.7+30=51.7s`