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

Answer:

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

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, $a=\frac{v-u}{t}$

                              $=\frac{39-0}{4.6}$

                              $= 8.478 \ m/s^2$

⇒ $F_{net}= Net \ force$

           $=ma$

           $=5\times 8.478$

           $=42.89 \ N$

Now,

The thrust will be:

⇒ $Thrust-weight-drag=Net \ force$

⇒ $Thrust-(5\times 9.8)-60=42.89$

⇒           $Thrust-49-60=42.89$

⇒                 $Thrust-109=42.89$

⇒                           $Thrust=42.89+109$

⇒                           $Thrust=151.89 \ N$

The height will be:

⇒ $h=\frac{1}{2} at^2$

      $=\frac{1}{2}\times 8.47\times (4.6)^2$

      $=\frac{1}{2}\times 8.47\times 21.16$

      $=89.61 \ m$