Question

During take-off a 5kg model rocket, initially at rest, burns fuel for 4.6s causing its speed to increase from rest to 39m/s during this time despite experiencing a 60N drag.

Answer 1

The height is "89.61 m". A further explanation is given below.

Step-by-step explanation:

According to the question,

Mass of rocket,

m = 5 kg

Time,

t = 4.6 s

Initial speed of rod

u = o m/s

Final speed,

v = 39 m/s

drag,

= 60 N

So,

⇒ acceleration,
`a=(v-u)/(t)`

`=(39-0)/(4.6)`

`= 8.478 \ m/s^2`

⇒
`F_(net)= Net \ force`

`=ma`

`=5* 8.478`

`=42.89 \ N`

Now,

The thrust will be:

⇒
`Thrust-weight-drag=Net \ force`

⇒
`Thrust-(5* 9.8)-60=42.89`

⇒
`Thrust-49-60=42.89`

⇒
`Thrust-109=42.89`

⇒
`Thrust=42.89+109`

⇒
`Thrust=151.89 \ N`

The height will be:

⇒
`h=(1)/(2) at^2`

`=(1)/(2)* 8.47* (4.6)^2`

`=(1)/(2)* 8.47* 21.16`

`=89.61 \ m`