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
$E_i = K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(0.181 kg)(92 m/s)^2 = 766 J$

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