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Aneli [31]
3 years ago
9

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.
Physics
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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Explanation:

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