Question

A rocket is launched with a constant thrust corresponding to an acceleration of ft/s2 . Ignoring air resistance, the rocket’s height after t second is given by (,) = 1 2 ( − 32) 2 ft. The limited fuel capacity of the rocket satisfies the equation 2 = 10000. Find the value of which maximizes the height that the rocket reaches when the fuel runs out. ​

Answer 1

Answer:

Relative velocity of exhaust gas V=2 km/s=2000 m/s

Thrust force F

t

=V

dt

dm

=2000×

dt

dm

Acceleration a=4.9 m/s

2

Mass of rocket M=1000 kg

Using Ma=F

t

−Mg

Or 1000×4.9=2000×

dt

dm

−1000×10

⟹

dt

dm

=7.35 kg/s

Step-by-step explanation:

Answer 2

Answer:

7.35 kg/s maximizes the height that the rocket reaches when the fuel runs out.

Step-by-step explanation:

As per the given data,

We need to find the value of which maximizes the height that the rocket reaches when the fuel runs out.

According to question,

It is given that,

A rocket is launched with a constant thrust corresponding to an acceleration of ft/s2 . Ignoring air resistance, the rocket’s height after t second is given by (,) = 1 2 ( − 32) 2 ft. The limited fuel capacity of the rocket satisfies the equation 2 = 10000.

Relative velocity of exhaust gas V=2 km/s=2000 m/s

Thrust force F

$t=v\frac{dm}{dt}$=2000×$\frac{dm}{dt}$

Acceleration a=4.9 m/s

Mass of rocket M=1000 kg

Using Ma=F

$\frac{-Mg}{t}$

Or 1000×4.9=2000×$\frac{dm}{dt}$−1000×10

⟹

$\frac{dm}{dt}$=7.35 kg/s

Hence,7.35 kg/s maximizes the height that the rocket reaches when the fuel runs out.

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