A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust ends. The rocket has mass 2 kg and thrust force 35 N. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, and (c) the speed of the rocket when it returns to the ground.


Answer 1


a) v=19.6 m/s

b) H=19.58 m

c) v_(f)=29.57 m/s  


a) Let's calculate the work done by the rocket until the thrust ends.

W=F_(tot)h=(F_(thrust)-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=(1)/(2)mv^(2)

v=\sqrt{(2W)/(m)}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.


At the maximum height the velocity is 0, so v(f) = 0.


H=(19.6^(2))/(2*9.81)=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)


v_(f)=\sqrt{19.6^(2)-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

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Young students show a preference for which modality?


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Be a qt today lolololololol



you too


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Salary compression

Occupational based pay

Merit pay

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In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.



Time period  = 41654.08 s


Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 * 10^(-5) kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s

total volume of air  infiltration= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s

Time period = (210)/(5.04* 10^(-3)) = 41654.08 s

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percentage yield = 63%


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If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K



0.25 J/K


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temperature (T) = 400 K

to find out

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ΔS = ΔQ / T           ...................a

Now we put the value of heat (Q) and Temperature (T) in equation a

ΔS is the entropy change, Q is heat and T is the temperature,  

so that

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ΔS = 0.25 J/K