15.291
m
1.766
s
17.321
m/s
10
m/s
10
m/s
15.291
m
15.291
m
1.766
s
17.321
m/s
10
m/s
10
m/s
15.291
m
The Maximum Height Calculator determines the peak altitude reached by a projectile launched at a given speed and angle. At maximum height, the vertical velocity component is zero while the horizontal component remains unchanged, making this a critical point in the trajectory analysis of any ballistic object.
The maximum height is derived from the kinematic equation $$v_y^2 = v_{0y}^2 - 2gH$$. Setting $$v_y = 0$$ at the apex yields: $$H = \frac{v_0^2 \sin^2(\theta)}{2g}$$ where $$v_0$$ is launch speed, $$\theta$$ is the launch angle, and $$g$$ is gravitational acceleration. The time to reach maximum height is half the total flight time: $$t_H = \frac{v_0 \sin(\theta)}{g}$$.
Maximum height depends on the square of the vertical velocity component ($$v_0 \sin\theta$$). A vertical launch ($$\theta = 90°$$) achieves the greatest possible height $$H_{\max} = v_0^2/(2g)$$ because all kinetic energy converts to gravitational potential energy. As the angle decreases, more energy goes into horizontal motion and less into vertical rise.
At the apex of the trajectory, the projectile has minimum speed but is not stationary (unless launched vertically). The speed at maximum height equals the horizontal component $$v_0 \cos\theta$$, which remains constant throughout the flight in the absence of air resistance. This is why a ball launched at 45° is moving at $$v_0/\sqrt{2}$$ at its highest point.
Understanding maximum height is essential in many practical scenarios: determining if a baseball clears an outfield wall, ensuring fireworks detonate at the correct altitude, calculating the apex of artillery shells for fragmentation timing, designing water fountain trajectories, and analyzing the biomechanics of jumping athletes.
The energy perspective provides additional insight: at maximum height, the kinetic energy $$\frac{1}{2}mv_0^2 \sin^2\theta$$ has converted to potential energy $$mgH$$, while the remaining kinetic energy $$\frac{1}{2}mv_0^2 \cos^2\theta$$ drives horizontal motion. This energy partition principle is fundamental to understanding all projectile trajectories.
Enter the initial velocity and launch angle. The calculator decomposes the velocity into vertical ($$v_0 \sin\theta$$) and horizontal ($$v_0 \cos\theta$$) components, then applies $$H = v_0^2 \sin^2\theta / (2g)$$ for maximum height and $$t_H = v_0 \sin\theta / g$$ for the time to reach it. Speed at maximum height equals the constant horizontal velocity component.
Maximum height increases with angle, reaching its absolute maximum at 90° (vertical launch). The time to maximum height is exactly half the total flight time. At the apex, the object moves purely horizontally at speed $$v_0 \cos\theta$$. Doubling initial speed quadruples the maximum height.
Inputs
Results
A ball at 20 m/s at 60° reaches 15.3 m high in 1.77 s. At the top, it moves horizontally at 10 m/s.
Inputs
Results
An arrow at 50 m/s at 75° reaches 119 m height—nearly the maximum possible 127.4 m at 90°.
At 90° (vertical launch), maximum height is $$v_0^2/(2g)$$. Any other angle reduces the vertical component and thus the peak altitude, since height depends on $$\sin^2\theta$$.
Only if launched vertically (90°). Otherwise, the horizontal velocity $$v_0 \cos\theta$$ persists at the apex. The vertical velocity is zero, making the speed equal to the horizontal component.
Time to maximum height is exactly half the total flight time for level ground: $$t_H = T/2 = v_0 \sin\theta / g$$. The ascent and descent durations are symmetric without air resistance.
Since $$H \propto v_0^2$$, doubling the speed quadruples the maximum height. This quadratic relationship makes initial speed the most influential factor in determining peak altitude.
The speed at the apex is $$v_0 \cos\theta$$—the horizontal velocity component, which remains constant throughout the flight. At 45°, this is about 71% of the launch speed.
Yes, significantly. Air resistance reduces maximum height by 10-40% for typical sports projectiles, and the effect increases with speed. The drag force opposes motion during both ascent and descent.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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