Question

A rocket is fired vertically and ascends with a constant vertical acceleration of 20 m/s2 for 1 min. Its fuel is then all used and it continues as a free fall particle. (a) Calculate the maximum altitude reached. (b) Calculate the total time elapsed from takeoff until the rocket strikes the earth.

Answer 1

Max height= 36000 metres

Total Time of flight = 120 sec

Step-by-step explanation:

It's acceleration is 20 m/s².

Time for vertical uplifting is one minute= 60 sec

V= U - at

At Max height, final velocity = zero

0 = U - at

U = at

U= 20*60

U= 1200 m/s

Formula for Max altitude

(U²Sin²tita)/2g

Max height= 1200² *( SIN90)²/(2*20)

Max height= 36000 metres

Time of flight= 2Usintita/g

= 2*1200/20

= 2400/20

= 120 sec

2 min

Answer 2

a)
`110\ \rm km`

b)
`330\ \rm s`

Step-by-step explanation:

Given:

• Acceleration of the rocket,
  `a=20\ \rm m/s^2`
• Time taken ,
  `t=60\ \rm s`

Equations of motion will be used to solve get maximum altitude and total time taken

Distance travelled by the rocket during acceleration

`s=(at^2)/(2)\\s=(20** 60^2)/(2)\\s=36000\ \rm m`

Let v be the velocity of the rocket at the end of 60 s which is given by

`v=at\\v=20*60\\v=1200\ \rm m/s`

a) Now the rocket will be under the gravity after its fuel ends, at the highest point of trajectory its final velocity will be zero

`0=1200-gt\\t=122\ \rm s`

During this time the rocket will rise to the maximum height under gravity

`y=36000+1200*122-(g*122^2)/(2)\\y=110*10^3\ \rm m`

Hence the maximum height of the rocket is calculated.

b) When the rocket falls back to the ground its displacement will be zero with respect to the ground

`0=3600+1200* t-(gt^2)/(2)\\t=270\ \rm s`

The total time
`=270 + 60=330\ \rm s`