Question

A model rocket is fired vertically from rest. It has a constant acceleration of 17.5m/s^2 for the first 1.5 s. Then its fuel is exhausted, and it is in free fall. (a) Ignoring air resistance, how high does the rocket travel? Cont.

(b) How long after liftoff does the rocket return to the ground?

Answer 1

(a) Given acceleration `a=15.5ms^-2` for first 1.5 seconds, after starting from rest. Let `v` be velocity attained when fuel is exhausted. Kinematic equation is
`v=u+a t` ........(1)

Inserting given values we get
`v=0+17.5xx1.5`
`=>v=27.25ms^-1` .....(2)

Using the following kinematic equation for finding height attained till `1.5s`
`v^2-u^2=2as`
`(27.25)^2-0^2=2xx17.5h`
`=>h=(27.25)^2/(2xx17.5)`
`=>h~~21.22m` ......(3)

These equations (2) and (3) give initial conditions for the freely falling rocket after fuel is exhausted.

Let rocket reach a maximum height `(h+h_1)` where velocity is zero. Acceleration due to gravity is in a direction opposite to the positive direction of motion.
To calculate height `h_1` attained under free fall we use the kinematic relation
`v^2-u^2=2as` ........(4)
`:.` `0^2-(27.25)^2=2xx(-9.81)h_1`
`=>h_1=(27.25)^2/19.62`
`=>h_1~~37.85m`

Maximum height attained is `h+h_1=21.22+37.85=59.07m`

(b) Let time taken to travel from height `h` to height `h_1` be `t_1`.
It can be found from the kinematic equation (1)
`0=27.25-9.81t_1`
`=>t_1=27.25/9.81=2.bar7s`

Now for downward journey of rocket, let the time taken for falling from maximum height to the ground be `t_2`. Applicable kinematic expression is
`s=ut+1/2at^2` ......(4)
Acceleration due to gravity is in the direction of motion. We have
`59.07=0xxt+1/2(9.81)t_2^2`
`=>t_2^2=sqrt((59.07xx2)/9.81)`
`=>t_2^2=approx3.47s`

Total time taken after liftoff`=1.5+t_1+t_2`
`=1.5+2.bar7+3.47=7.7s`, rounded to one decimal place

-.-.-.-.-.-.-.-.-.-.-.

Alternate method for part (b)
After `1.5s`. Rocket is falling freely under gravity. When it reaches ground height `=0`.
Displacement`="Final position" -"Initial position"`
`=0-21.22=-21.22m`
Time taken to reach ground can be calculated using (4). Acceleration due to gravity acting against the direction of motion.
`-21.22=27.25t+1/2(-9.81)t^2`
`=>9.81t^2-54.5t-42.44=0`
Roots of this quadratic can be found using

`t=(-b+-sqrt(b^2-4ac))/(2a)`

Using inbuilt graphic tool.

Ignoring the negative root as time can not be negative. we have
Time of flight `t` after fuel is consumed `=6.2s`, rounded to one decimal place.
Total time taken after liftoff`=1.5+t`
`=1.5+6.2=7.7s`, rounded to one decimal place