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?