In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is why that when I toss an object directly upwards, the kinetic energy $K = (1/2)mv^2$ is transformed into potential energy as it increases in height with potential energy $U = mgh$

Because of the conservation due to the energies being transformed, we can express this relationship between the two energies as $K_i + U_i = K_f + U_f$

The question I was asked was to use these equations to find the maximum height $h_{max}$ to which the object will rise, as expressed in terms in $v$ and $g$.

I was able to solve this by saying that at this max height, the velocity and therefore the kinetic energy will be at zero. So I am able to say that $K_i + 0 = 0 + U_f$ or simply $K_i = U_f$

Mass cancels and we are left with $\frac{v^2}{2g} = h_{max}$

This was easy enough, and it is likely that my misunderstanding is simply a mathematical one, but I am at a loss when asked

At what height $h$ above the ground does the projectile have a speed of $0.5v$

How do you approach this problem?


While writing out my progress on the problem, I managed to give myself the answer. So, I thought that I may as well share the solution as I have seen many people in my class get stuck here.

If I have a kinetic energy equal to $K = (1/2)mv^2$

And I later have a velocity equal to half the original $v$

What happens to $K$? Shouldn't it be 1/4th the original? Plugging in numbers shows that this is indeed the case.

So if my $K$ is 1/4th what it originally was, my $U$ should now be 3/4ths of my original $K$

Write that out using $K_i + U_i = K_f + U_f$

$K_i = \frac{1}{4}K_i + U_f$

$\frac{3}{4}K_i = U_f$

so the question, again, was at what height $h$ above the ground is the speed $\frac{1}{2}v$

The answer is $h = \frac{3v^2}{8g}$

You simply plug in the standard $K$ and $U$ equations into $\frac{3}{4}K_i = U_f$ and solve for $h$ as usual.

  • $\begingroup$ There's no need to repeat all your algebra. Once you find that U is 3/4 the original K, the height is 3/4 the max height you found in the previous problem. $\endgroup$ – Mark Eichenlaub Dec 5 '12 at 9:56

This isn't an answer, but rather a note on Leonardo's working (which is entirely correct).

Suppose your initial velocity is $u$, and you're asking what height corresponds to what velocity $v$. This is a generalised version of Leonardo's question for which $v$ is $u/2$. Anyhow equating energy just like Leonardo does we get.

$$ \frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgs $$

where $g$ is the acceleration due to gravity and $s$ is the distance travelled. Rearranging this gives:

$$ v^2 = u^2 - 2gs $$

Now the acceleration due to gravity is in the opposite direction to the velocities, so if we take the sign convention that upwards is positive (to match the velocities) we should write the acceleration as $a = -g$, and putting this into the equation gives:

$$ v^2 = u^2 + 2as $$

And this is one of the standard equations of motion we learned at school. So, Leonardo, not only have you answered your specific question, but (maybe) without realising it you've actually derived the equation of motion for the particle throughout it's whole path! :-) In fact you've found one of the SUVAT equations.


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