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3 years ago

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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 duri

ng this time despite experiencing a 60N drag.

1 answer:

MrRa [10]3 years ago

5 0

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:

⇒ $h=\frac{1}{2} at^2$

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

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

$=89.61 \ m$

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