KAHIBARO
Table of Contents
Reaching the Top of a Projectile's Motion
Maximum height is the highest vertical position reached by a projectile during its flight. This idea is specific to the upward part of the motion. At the top point, the projectile stops rising for an instant before it starts falling.
The key fact is that the vertical component of velocity becomes zero at the highest point. The projectile may still have horizontal motion, but vertically it is momentarily at rest.
At maximum height, the vertical velocity is zero:
$$v_y = 0$$
This does not mean the total velocity is zero, unless there is no horizontal motion.
Vertical Motion Determines the Maximum Height
To find the maximum height, we look only at the vertical motion. The horizontal motion does not affect how high the projectile rises.
If the projectile is launched with initial speed $v_0$ at an angle $\theta$ above the horizontal, then its initial vertical velocity is
$$v_{0y} = v_0 \sin\theta$$
As the projectile rises, gravity slows this vertical motion with acceleration
$$a_y = -g$$
Using the kinematic relation for vertical motion,
$$v_y^2 = v_{0y}^2 + 2a_y(y - y_0)$$
and setting $v_y = 0$ at the top, we get
$$0 = v_{0y}^2 - 2g(y_{\max} - y_0)$$
So the vertical rise above the launch point is
$$y_{\max} - y_0 = \frac{v_{0y}^2}{2g}$$
Substituting $v_{0y} = v_0\sin\theta$ gives
$$y_{\max} - y_0 = \frac{v_0^2 \sin^2\theta}{2g}$$
Maximum height above the launch point:
$$H = \frac{v_{0y}^2}{2g} = \frac{v_0^2 \sin^2\theta}{2g}$$
Meaning of the Formula
This formula shows that the maximum height depends on the initial vertical speed, not directly on the horizontal speed. If two projectiles have the same initial vertical component of velocity, they reach the same maximum height, even if their horizontal speeds are different.
A larger launch angle usually gives a larger vertical component, as long as the launch speed stays the same. That means the projectile rises higher.
The gravitational acceleration $g$ appears in the denominator, so a stronger gravitational field gives a smaller maximum height.
If the Launch Point Is Not Ground Level
Sometimes the projectile is launched from an initial height $y_0$ above the ground. Then the highest vertical position measured from the ground is
$$y_{\max} = y_0 + \frac{v_0^2 \sin^2\theta}{2g}$$
This expression gives the actual top position, not just the rise above the launch point.
Visualizing the Highest Point
The highest point is where the projectile changes from upward motion to downward motion. The path is curved, but the top is easy to identify because the slope of the path becomes horizontal there.
Useful Special Case
If the projectile is thrown straight upward, then $\theta = 90^\circ$ and $\sin\theta = 1$. The formula becomes
$$H = \frac{v_0^2}{2g}$$
This is the greatest possible height for a given launch speed, because all of the initial speed is vertical.
Common Mistakes
A common mistake is to set the total velocity equal to zero at the top. That is only true for purely vertical motion. In ordinary projectile motion, only the vertical component is zero.
Another common mistake is to forget whether the formula gives height above the launch point or height above the ground. The quantity
$$\frac{v_0^2 \sin^2\theta}{2g}$$
is the rise relative to where the projectile started.
Be careful:
$$\frac{v_0^2 \sin^2\theta}{2g}$$
is the increase in height above the launch point, not always the total height above the ground.
Summary Table
| Quantity | Expression |
|---|---|
| Initial vertical velocity | $v_{0y} = v_0\sin\theta$ |
| Vertical velocity at top | $v_y = 0$ |
| Maximum height above launch point | $H = \dfrac{v_0^2\sin^2\theta}{2g}$ |
| Maximum vertical position from ground | $y_{\max} = y_0 + \dfrac{v_0^2\sin^2\theta}{2g}$ |
Final Idea
Maximum height is found by focusing on the vertical motion and using the fact that the vertical velocity is zero at the top. This makes the highest point one of the simplest and most important features of projectile motion.