Question

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`

Step-by-step explanation:

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.

`v_(f)^(2)=v_(i)^(2)-2gh`

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

`0=v_(i)^(2)-2gH`

`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)^(2)=v_(i)^(2)-2gh`

`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!