Energy Conservation for Launch Problems

When an object is launched as a projectile, mechanical energy is conserved if air resistance is negligible. The total energy $E = K + U$ remains constant, transforming between kinetic and gravitational potential energy.

Object thrown straight up:

Consider an object of mass $m$ thrown upward from ground level with initial speed $v_0$.

Initially (ground level), where $U_i = 0$:

$$E_i = \frac{1}{2}mv_0^2 + U_i = \frac{1}{2}mv_0^2$$

At maximum height (velocity zero), where $K_f = 0$:

$$E_f = K_f + mgh_{\text{max}} = mgh_{\text{max}}$$

Conservation gives:

$$\frac{1}{2}mv_0^2 = mgh_{\text{max}}$$

Mass cancels:

$$h_{\text{max}} = \frac{v_0^2}{2g}$$

Note we obtained this formula without using any kinematic equations.

Speed at any height:

At height $h$, the object has both kinetic and potential energy.

Applying energy conservation:

$$E_i = E_f$$

Initially (at ground), where $U_i = 0$:

$$K_i + U_i = \frac{1}{2}mv_0^2$$

At height $h$:

$$K_f + U_f = \frac{1}{2}mv^2 + mgh$$

Conservation gives:

$$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$$

Mass $m$ cancels:

$$\frac{1}{2}v_0^2 = \frac{1}{2}v^2 + gh$$

Solving for $v$:

$$v = \sqrt{v_0^2 - 2gh}$$

This applies during ascent or descent—the speed at height $h$ is the same going up or down.

Return to ground:

When the object returns to ground level, the potential energy is again zero.