Question

A model rocket with a mass of 0.181 kg is launched into the air with an initial speed of 92 m/s. How much kinetic energy will the rocket have at a height of 164 m? Assume there is no wind resistance. 634 J 444 J 747 J 303 J

Answer 1

We can solve the problem by using conservation of energy.

Initially, the total energy of the rocket is only kinetic energy. This is equal to

`E_i = K_i = (1)/(2) m v_i^2 = (1)/(2)(0.181 kg)(92 m/s)^2 = 766 J`

As the rocket goes higher, part of this kinetic energy converts into potential energy. Indeed, at the height h=164 m, the total energy of the rocket is the sum of the kinetic energy (at the new speed
`v_f`) and the potential energy:

`E_f = K_f + E_p = K_f + mgh`

For the conservation of energy,
`E_i = E_f`, so we can write:

`K_i = K_f + mgh`
and so we can find the kinetic energy at height h=164 m:

`K_f = K_i - mgh = 766 J-(0.181 kg)(9.81 m/s^2)(164 m)=475 J`