Interactive physics simulator
Maximum Height of Projectile
Explore how the initial launch speed, launch angle, and gravity determine the absolute peak height a projectile can reach. Watch the vertical speed drop to zero at the apex while horizontal motion carries it forward.
Maximum Height Simulator
Configure velocity, launch angle, and gravity environment. Observe the vertical trajectory and find how they scale the apex height.
Live Result
- Initial Speed (v_0)
- 15 m/s
- Launch Angle (θ)
- 45°
- Initial Vertical Speed (v_0y)
- 10.6 m/s
- Initial Horizontal Speed (v_0x)
- 10.6 m/s
- Vertical Speed (v_y)
- 10.6 m/s
- Horizontal Speed (v_x)
- 10.6 m/s
- Time Elapsed (t)
- 0 s
- Current Height (y)
- 0 m
- Maximum Height (H)
- 5.74 m
- Time to Peak (t_apex)
- 1.08 s
- Gravity (g)
- 9.8 m/s²
- Apex Formula
- H = v_0y² / 2g
What is the Maximum Height of a Projectile?
The maximum height (denoted by H) is the highest vertical position reached by a projectile during its flight. When an object is launched upward at an angle, gravity decelerates its vertical velocity component vy until it momentarily becomes exactly 0 m/s at the peak (apex) of its trajectory.
Although the vertical velocity is zero at this point, the projectile is not stationary; its constant horizontal velocity vx continues to carry it forward. After passing this peak height, the vertical velocity reverses and the object descends.
Key Ideas of Maximum Height
Understanding the physics at the apex of projectile flight.
- Vertical Velocity is Zero: At the highest point, vy = 0 m/s.
- Constant Horizontal Velocity: The horizontal speed vx stays constant and equal to v0 cos(θ).
- Constant Acceleration: The acceleration remains constant at g (9.8 m/s² downward) even at the very top.
- Independence of Motion: The peak height depends only on the vertical launch velocity component and gravity.
Kinematic Formulas
H = (v0 × sin(θ))2 / 2g
v0y = v0 × sin(θ)
tapex = v0 × sin(θ) / g
Important derived relations:
- Maximum height in terms of initial vertical velocity: H = v0y2 / 2g
- Time to peak is half of the total level-ground time of flight.
- Straight vertical launch (θ = 90°) yields: H = v02 / 2g
Factors Affecting Height
- Launch Speed (v0): Maximum height scales with the square of launch speed (H ∝ v02). Doubling speed quadruples the maximum height.
- Launch Angle (θ): Height depends on sin2(θ). Peak height is maximized at 90° and is exactly zero for a horizontal launch (0°).
- Gravity (g): Height is inversely proportional to gravity (H ∝ 1/g). In weaker gravity (e.g. Moon), projectiles rise much higher.
Visual Summary
A ball launched at 15 m/s at a 45° angle reaches its maximum height. Here is how the trajectory and velocity look at the peak:
Solved Examples
A baseball is hit with an initial speed of 30 m/s at an angle of 35° above the horizontal. Calculate the maximum height it reaches. Use g = 9.8 m/s².
- Identify initial values: v0 = 30 m/s, launch angle θ = 35°, gravity g = 9.8 m/s².
- Calculate the initial vertical velocity component: v0y = v0 × sin(θ).
- v0y = 30 × sin(35°) ≈ 30 × 0.5736 ≈ 17.21 m/s.
- Use the maximum height formula: H = v0y2 / (2g).
- H = (17.21)2 / (2 × 9.8) = 296.18 / 19.6 ≈ 15.11 meters.
Answer: Maximum Height ≈ 15.11 m
An athlete performs a long jump, launching at an angle of 22° and reaching a peak height of 0.85 m. What was their initial launch speed? Use g = 9.8 m/s².
- Given launch angle θ = 22°, maximum height H = 0.85 m, and gravity g = 9.8 m/s².
- Rearrange the maximum height formula H = (v0 × sin(θ))2 / (2g) to solve for v0:
- v0 = √(2gH) / sin(θ).
- First calculate the vertical speed at launch: v0y = √(2 × 9.8 × 0.85) = √(16.66) ≈ 4.08 m/s.
- Now find the total launch speed: v0 = 4.08 / sin(22°) = 4.08 / 0.3746 ≈ 10.89 m/s.
Answer: Initial speed ≈ 10.89 m/s
A water balloon launcher fires a balloon on Mars (g = 3.71 m/s²) with a speed of 18 m/s at a 60° angle. Find the maximum height reached.
- Identify known values on Mars: v0 = 18 m/s, θ = 60°, g = 3.71 m/s².
- Calculate the vertical component: v0y = 18 × sin(60°) = 18 × 0.8660 = 15.59 m/s.
- Calculate maximum height: H = v0y2 / (2g).
- H = (15.59)2 / (2 × 3.71) = 243.05 / 7.42 ≈ 32.76 meters.
- Note: Due to the lower gravity on Mars, the balloon reaches more than double the height it would reach on Earth (≈ 12.4 m).
Answer: Maximum Height ≈ 32.76 m
Common Mistakes
- Using total launch velocity in height calculation: Using v0 instead of the vertical component v0y = v0sin(θ). The horizontal part carries the object forward but does not lift it.
- Thinking total velocity is zero at peak: Assuming that the projectile stops completely at its highest point. At the peak, only vertical velocity is zero. Horizontal velocity vx remains constant and is non-zero.
- Confusing equations of motion: Misapplying formulas designed for horizontal projectile launches where initial v0y is 0.
- Forgetting to square the sine component: Writing (v02 × sin(θ)) instead of (v0 × sin(θ))2. The entire numerator term must be squared.
Quick Summary
- Maximum height occurs when the vertical velocity of the projectile reaches exactly 0 m/s.
- Formula: H = v0y2 / 2g = (v0 × sin(θ))2 / 2g.
- Time to maximum height is half of the total level-ground flight time: tapex = v0y / g.
- Maximum height is directly proportional to the square of initial speed and launch angle sine component, and inversely proportional to gravity.
- Mass has zero effect on the maximum height in a vacuum model.
Practice Questions
1. A football is kicked with a velocity of 22 m/s at an angle of 40°. Find its maximum height. (Use g = 9.8 m/s²)
v0y = 22 × sin(40°) ≈ 14.14 m/s. H = v0y2 / 2g = (14.14)2 / (2 × 9.8) ≈ 10.20 meters.
2. An arrow is launched vertically upward (θ = 90°) at 25 m/s. What is its maximum height? (Use g = 9.8 m/s²)
Since θ = 90°, sin(θ) = 1. The formula reduces to H = v02 / 2g = 252 / (2 × 9.8) = 625 / 19.6 ≈ 31.89 meters.
3. How does the maximum height of a projectile change if its initial speed is doubled while the launch angle remains the same?
Maximum height is proportional to the square of initial speed (H ∝ v02). Therefore, doubling the initial speed increases the maximum height by a factor of 4 (22).
4. If a projectile is launched at 45° and another at 60° with the same initial speed, what is the ratio of their maximum heights?
The maximum height is proportional to sin2(θ). The ratio is sin2(45°) / sin2(60°) = (1/√2)2 / (√3/2)2 = 0.5 / 0.75 = 2/3 (or 1 : 1.5).
5. At the peak of a projectile's trajectory, what are its vertical and horizontal velocity components?
The vertical velocity component (vy) is exactly 0 m/s. The horizontal velocity component (vx) is constant and equal to its initial value (v0 × cos(θ)).
6. A launcher on the Moon (g = 1.62 m/s²) shoots a projectile straight up at 12 m/s. What is the peak height reached?
H = v02 / 2g = 122 / (2 × 1.62) = 144 / 3.24 ≈ 44.44 meters.
FAQ
Frequently Asked Questions
What is the maximum height of a projectile?
The maximum height (H) is the highest vertical position reached by a projectile during its flight. At this exact point, the projectile's vertical velocity component becomes momentarily zero before it begins its descent.
What is the formula for maximum height?
The formula for maximum height is: H = (v0 × sin(θ))2 / (2g), where v0 is the initial launch speed, θ is the launch angle above the horizontal, and g is the acceleration due to gravity.
Why does vertical velocity become zero at maximum height?
Gravity exerts a constant downward force on the projectile. As it rises, gravity decelerates its upward motion at a rate of g (9.8 m/s² on Earth) until the vertical velocity reaches zero. At that instant, it stops rising and starts falling.
Does horizontal velocity affect the maximum height?
No. Horizontal and vertical motions are independent. Horizontal velocity remains constant (ignoring air resistance) and determines how far the projectile travels horizontally, but only vertical velocity determines the peak height.
What launch angle yields the absolute maximum height?
A launch angle of 90° (straight vertical launch) yields the maximum height for any given initial speed, because all of the launch energy is directed vertically (sin(90°) = 1).
How does gravity affect maximum height?
Maximum height is inversely proportional to gravity (H ∝ 1/g). If gravity is weaker (like on the Moon where g ≈ 1.62 m/s²), the maximum height reached will be significantly higher than on Earth for the same launch speed and angle.
Is acceleration zero at the maximum height of flight?
No. Even though the vertical speed is momentarily zero at the apex, gravity continues to pull the object downward. Thus, the acceleration remains constant at g (9.8 m/s² downward) throughout the entire flight.
Does mass affect the maximum height of a projectile?
In an ideal vacuum model where air resistance is ignored, mass does not affect the trajectory. All objects, regardless of mass, experience the same gravitational acceleration and reach the same maximum height.
What is the relationship between launch speed and maximum height?
Maximum height is proportional to the square of the launch speed (H ∝ v02). For example, if you triple the launch velocity, the maximum height increases by a factor of 9.
Can a projectile have a negative maximum height?
No. By definition, height is measured upward from the launch position. A projectile launched horizontally (0°) or downward from a platform will simply descend, meaning its launch height is its highest point.
How do you find the time it takes to reach maximum height?
The time to reach the apex (tapex) is calculated by dividing the initial vertical velocity by gravity: tapex = (v0 × sin(θ)) / g.
What is the vertical velocity of a projectile just after reaching maximum height?
Immediately after reaching the apex, the vertical velocity becomes negative (directed downward) and increases in magnitude as the projectile accelerates toward the ground due to gravity.