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?
Best Answer
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.