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Degger [83]
3 years ago
12

A 10000 kg rocket blasts off vertically from the launch pad with a constant upward of 2.25 m/s2 and feels no appreciable air res

istance. When it has reached a height of 525 m, its engine suddenly fails so that only gravity acts on it after that. What is the maximum height that the rocket reaches?
Physics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

v^2=u^2+2as

Put the value in the equation

v^2=0+2\time2.25\times525

v=\sqrt{2\times2.25\times525}

v=48.60\ m/s

We need to calculate the maximum height with initial velocity

Using equation of motion

v^2=u^2-2gh

h=\dfrac{v^2-u^2}{-2g}

Put the value in the equation

h=\dfrac{0-(48.60)^2}{-2\times9.8}

h=120.50\ m

The total height  reached by the rocket is

h'=s+h

h'=525+120.50

h'=645.5\ m

Hence, The maximum height that the rocket reaches is 645.5 m.

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Answer:

a) t=24s

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Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

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w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

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