The following is the problem that I am working on.

Three different objects each with mass $m_1<m_2<m_3$ is launched from the same height $h$ with three different angles $\theta_1 < \theta_2 < \theta_3$. Each object has the same initial kinetic energy, $k$. Which object has the greatest speed just as it impacts with the ground ?

This is my claim.

The total mechanical energy of the three objects, say, $M_1,M_2$ and $M_3$ can be calculated as $$M_1 = K+m_1gh$$ $$M_2=K+m_2gh$$ $$M_3=K+m_3gh$$

Since the initial kinetic energy is the same for all three, the one with the largest mass, $m_3$ has the largest mechanical energy.

So I am thinking that since the height equals $0$ at the moment of impact, and all potential energy is converted into kinetic energy, the one with the largest total mechanical energy is the one with the fastest speed.

Yet, the answer says that it is the one with the smallest mass is the one with the greatest speed.

Can someone explain me this situation ?


1 Answer 1


As long as the $\angle \theta_n$ for each is $0^\circ < \angle \theta_n < 180^\circ$ then their $\angle \theta_n$ and trajectories don't really matter. They will start at at the same $h_0, v =v_0$, travel to some $h_\mathrm{max}, v_\mathrm{max}=0$ and then fall back down to $h$ traveling the same speed $v = v_0$. At this point they continue to fall from $h$ to $0$ gaining additional kinetic energy as their potential energy at $h$ is converted to additional kinetic energy.

Since all objects fall (accelerate) at the same speed in a gravitational field, whichever started the fastest at $h$ will hit the fastest at $0$.

Since all three started with the same kinetic energy but different masses, the one with the smallest mass must have had the greatest initial velocity $v_0$ and therefor will have the greatest velocity when it hits the ground.

  • $\begingroup$ Another way to say it is that since they start and end at the same height, their total energy at the beginning and the end is the same and given by $K$ (because energy is conserved). $K=\frac{1}{2}mv^2$, so the one with the lowest mass must have the highest velocity to make its kinetic energy equal to the heavier one's. $\endgroup$
    – Geoffrey
    Commented Dec 21, 2013 at 4:27
  • 1
    $\begingroup$ Actually, thinking about this more, the initial angle doesn't matter at all. Not even the restriction I mentioned. $\endgroup$ Commented Dec 21, 2013 at 4:34

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