Source: High school physics (Chinese)
Problem
A body is in vertical projectile motion under gravity, neglecting air resistance.
[Q1] The proof relies on the conservation of mechanical energy ($E = K + U_g$). At any given height $y$, the potential energy $U_g = mgy$ is the same for both ascent and descent. Since total energy $E$ is conserved, the kinetic energy $K = \frac{1}{2}mv^2$ must also be the same.
- Equal Speeds: From $\frac{1}{2}mv_{up}^2 = \frac{1}{2}mv_{down}^2$, it follows that the speeds are identical: $$|v_{up}| = |v_{down}|$$
- Opposite Velocities: Since the direction of motion during ascent is opposite to that during descent, the velocities are equal in magnitude but opposite in direction: $$v_{down} = -v_{up}$$
We analyze the motion using the principle of conservation of mechanical energy. As the body moves under gravity without air resistance, its total mechanical energy, which is the sum of its kinetic energy ($K$) and gravitational potential energy ($U_g$), remains constant.
Let's consider an arbitrary point in the trajectory at a height $y$ above a chosen reference level ($y=0$). Let $v_{up}$ be the velocity of the body when it passes this point during its ascent (moving upwards). Let $v_{down}$ be the velocity of the body when it passes the same point during its descent (moving downwards).
The kinetic energy is $K = \frac{1}{2}mv^2$ and the gravitational potential energy is $U_g = mgy$.
The total mechanical energy $E$ at height $y$ during the ascent is:
$$E_{ascent} = K_{up} + U_g = \frac{1}{2}mv_{up}^2 + mgy$$The total mechanical energy $E$ at the same height $y$ during the descent is:
$$E_{descent} = K_{down} + U_g = \frac{1}{2}mv_{down}^2 + mgy$$According to the principle of conservation of energy, the total energy must be the same at all points in the trajectory. Therefore, $E_{ascent} = E_{descent}$.
$$\frac{1}{2}mv_{up}^2 + mgy = \frac{1}{2}mv_{down}^2 + mgy$$Subtracting the common potential energy term $mgy$ from both sides, we get:
$$\frac{1}{2}mv_{up}^2 = \frac{1}{2}mv_{down}^2$$Canceling the term $\frac{1}{2}m$ (as mass $m$ is non-zero):
$$v_{up}^2 = v_{down}^2$$Taking the square root of both sides gives the relationship for the magnitudes of the velocities, which are the speeds:
$$|v_{up}| = |v_{down}|$$This proves that the speed at any specific height is the same during ascent and descent.
Regarding the velocities, velocity is a vector quantity that includes direction. By definition, during ascent, the body is moving in the upward direction, while during descent, it is moving in the downward direction. Since the motions are in opposite directions and the magnitudes (speeds) are equal, the velocities are opposites of each other. If we define the upward direction as positive, then $v_{up}$ is positive and $v_{down}$ is negative. Thus, we can write the vector relationship as:
$$v_{down} = -v_{up}$$This demonstrates that the velocities at the same point are equal in magnitude and opposite in direction.