Interactive physics simulator
Vertical Projectile Motion
Study the physics of objects thrown straight up or down. Real-time graphs and vector representations demonstrate how gravity acts as a constant downward acceleration, decelerating upward motion and accelerating downward fall.
Vertical Projectile Simulator
Configure height, launch speed, and gravity. See the height and velocity change in real-time alongside synced graphs.
Live Result
- Initial Speed (v_0)
- 15 m/s
- Launch Height (h)
- 10 m
- Current Height (y)
- 10 m
- Current Velocity (v)
- 15 m/s
- Time Elapsed (t)
- 0 s
- Gravity (g)
- 9.8 m/s²
- Apex Height (H_max)
- 21.5 m
- Time to Apex
- 1.53 s
- Time of Flight
- 3.61 s
- Impact Velocity
- -20.4 m/s
- Direction
- Upward
- Active Equation
- y = h + v_0t - 1/2gt²
What is Vertical Projectile Motion?
Vertical projectile motion refers to the straight line, one-dimensional vertical motion of an object launched directly upward or downward. Once launched, it moves under the sole influence of gravity (with air resistance ignored).
The key characteristics of this motion are:
- Symmetry: When launched upward from the ground, the time it takes to go up to its apex is exactly equal to the time it takes to fall back down to the ground. The object also lands with the exact same speed as it was launched.
- Acceleration: Acceleration is constant and directed downwards throughout the entire flight (*ay = -g*). Even at the highest point of its flight where the object's velocity is momentarily zero, gravity continues to accelerate it at 9.8 m/s² downward.
Key Ideas of Vertical Launch
Acceleration remains constant throughout, even when the object is momentarily at rest at its highest point.
- Initial Velocity Direction: A positive initial velocity (v0 > 0) indicates an upward launch. A negative initial velocity (v0 < 0) indicates a downward launch.
- The Apex Point: At the highest point of an upward trajectory, velocity becomes exactly 0 m/s. Acceleration is still *g* downward.
- Constant Gravity: Gravitational acceleration does not turn off at the top of the flight.
- Air Resistance: In the ideal model, objects of different weights fall at the exact same rate.
Kinematic Formulas
v = v0 - gt
y = h + v0t - ½gt2
v2 = v02 - 2g(y - h)
Derived Apex & Flight equations:
- Time to Apex: tapex = v0 / g (if launched upward)
- Apex Height: Hmax = h + v02 / (2g)
- Time of Flight (to ground y=0): Solve ½gt2 - v0t - h = 0
- Impact Speed: v = √(v02 + 2gh)
Motion States Breakdown
| Flight Stage | Velocity (v) | Acceleration (a) | Direction of Motion |
|---|---|---|---|
| Going Up | Positive (decreasing) | -g (downward) | Upward |
| At Apex | 0 m/s | -g (downward) | Momentarily Stationary |
| Going Down | Negative (increasing magnitude) | -g (downward) | Downward |
| Impact | Maximum Negative Speed | -g (downward) | Downward |
Real-life Example
A camera is thrown straight up from a 15 m high balcony with an initial speed of 10 m/s.
Initial Velocity: v0 = 10 m/s. Elevation: h = 15 m.
Calculating the maximum height reached:
Hmax = 15 + 102 / (2 × 9.8) = 15 + 100/19.6 ≈ 20.1 m.
Time to reach that highest point:
tapex = 10 / 9.8 ≈ 1.02 s.
Solved Examples
A ball is thrown vertically upward from the ground with an initial velocity of 20 m/s. Find the maximum height reached and the time taken to reach that height. Use g = 9.8 m/s².
- Identify initial values: launch height h = 0 m, initial upward velocity v0 = 20 m/s, gravity g = 9.8 m/s².
- At the highest point (apex), vertical velocity v = 0 m/s.
- Use the velocity formula: v = v0 - gt. Setting v = 0, we get tapex = v0 / g.
- tapex = 20 m/s / 9.8 m/s² ≈ 2.04 s.
- Use the maximum height formula: Hmax = h + v02 / (2g).
- Hmax = 0 + 202 / (2 × 9.8) = 400 / 19.6 ≈ 20.4 m.
Answer: Time to apex = 2.04 s, Maximum Height = 20.4 m
A stone is thrown vertically downward from a bridge 30 m high with an initial speed of 5 m/s. Calculate the time it takes to hit the water below and its impact velocity. Use g = 9.8 m/s².
- Let upward be positive. Launch height h = 30 m, initial velocity v0 = -5 m/s (negative since it is downward), g = 9.8 m/s².
- Calculate impact speed using the formula: v2 = v02 - 2g(y - h). At y = 0, v2 = v02 + 2gh.
- v = √((-5)2 + 2 × 9.8 × 30) = √(25 + 588) = √613 ≈ 24.8 m/s (downward, so v = -24.8 m/s).
- Use the first equation of motion to find time: v = v0 - gt.
- -24.8 = -5 - 9.8t ⇒ 9.8t = 19.8 ⇒ t ≈ 2.02 s.
Answer: Time = 2.02 s, Impact velocity = 24.8 m/s downward
A model rocket is launched vertically upward from a 10 m high platform with an initial velocity of 30 m/s. Find its height and velocity after 4.0 seconds of flight. Use g = 9.8 m/s².
- Identify known values: launch height h = 10 m, initial upward velocity v0 = 30 m/s, time t = 4.0 s, g = 9.8 m/s².
- Calculate vertical position: y = h + v0t - ½gt2.
- y = 10 + (30 × 4) - ½(9.8 × 42) = 10 + 120 - 4.9 × 16 = 130 - 78.4 = 51.6 m.
- Calculate velocity: v = v0 - gt.
- v = 30 - (9.8 × 4) = 30 - 39.2 = -9.2 m/s (the negative sign indicates the rocket is moving downward).
Answer: Height = 51.6 m, Velocity = 9.2 m/s downward
Common Mistakes
- Thinking acceleration is zero at the apex: Assuming that because the projectile is momentarily stopped, gravity has stopped pulling. Gravity is constant, so acceleration is always 9.8 m/s² downward.
- Sign errors: Forgetting that downward velocity and gravity have opposite signs to upward velocity (e.g. adding gravity instead of subtracting it when computing upward flight).
- Mass assumption: Thinking a heavier object falls faster or reaches a lower apex height than a lighter one. Mass does not affect ideal kinematic calculations.
- Forgetting launch height: Neglecting the initial height *h* when calculating position *y*, or when finding impact speed.
Quick Summary
- Vertical launch angle is exactly 90° (upward) or -90° (downward).
- Acceleration is constant: a = -g (downward).
- Velocity changes linearly: v = v0 - gt.
- Position changes quadratically: y = h + v0t - ½gt2.
- Velocity at the apex of flight is exactly 0 m/s.
- Motion is symmetrical when launched and landed on the same level.
Practice Questions
1. A tennis ball is thrown straight up into the air at 15 m/s. How long does it stay in the air before landing back on the ground? (Use g = 9.8 m/s²)
For level ground, total time of flight T = 2v0 / g = 2 × 15 / 9.8 ≈ 3.06 s.
2. An object is dropped from a window 19.6 m high. What is its velocity right before hitting the ground? (Use g = 9.8 m/s²)
v = √(2gh) = √(2 × 9.8 × 19.6) = √(384.16) = 19.6 m/s downward.
3. If a ball is thrown vertically upward with speed 25 m/s on Mars where g = 3.71 m/s², what is the maximum height reached?
Hmax = v02 / 2g = 252 / (2 × 3.71) = 625 / 7.42 ≈ 84.2 meters.
4. What is the vertical acceleration of a ball thrown straight up at the highest point of its flight?
It is exactly 9.8 m/s² downward (gravity). Even though velocity is zero, gravity pulls constant, keeping acceleration at g.
5. You throw a key straight up at 8 m/s from a 12 m high window. What is its speed when it passes the window on the way back down?
Ignoring air resistance, vertical motion is symmetric. It passes the window downward at exactly the same speed it was thrown upward (8 m/s downward).
6. If initial vertical velocity is doubled, what happens to the maximum height reached?
Hmax is proportional to v02. If you double initial velocity, the maximum height increases by a factor of 4 (22).
FAQ
Frequently Asked Questions
What is vertical projectile motion?
Vertical projectile motion is a type of one-dimensional motion where an object is launched straight upward or straight downward, moving solely under the influence of gravity with air resistance ignored.
What is the acceleration of a vertical projectile at its highest point?
The acceleration is exactly g (9.8 m/s² downward on Earth). Even though the velocity of the projectile is momentarily zero at the highest point, gravity is still pulling it downward, so the acceleration remains constant.
How does launch height affect the time of flight and impact speed?
A higher launch platform gives the projectile more vertical distance to fall, which increases both the total time of flight and the final speed at which it impacts the ground.
What is the formula for the maximum height reached by a vertical projectile launched upward?
The maximum height from the ground is given by: Hmax = h + v02 / (2g), where h is the launch height, v0 is the initial upward speed, and g is gravity.
How do you calculate the time of flight for a vertical projectile?
The time of flight is found by solving the quadratic position equation for y(t) = 0: ½gt2 - v0t - h = 0. This yields t = (v0 + √(v02 + 2gh)) / g (taking the positive root).
What is the difference between launching an object vertically upward versus launching it vertically downward?
An upward launch starts with a positive initial velocity, travels upward to an apex, stops, and then falls. A downward launch starts with a negative initial velocity and immediately speeds up downward, hitting the ground much faster with a shorter time of flight.
Does the mass of the projectile affect its vertical motion?
In the absence of air resistance, all objects fall at the same rate regardless of their mass. Mass does not appear in any of the kinematics equations for vertical projectile motion.
Why is the Position-Time graph of a vertical projectile curved?
Because gravity causes constant downward acceleration, the vertical position changes quadratically with time (y ∝ t²), which creates a parabolic curve on a Position-Time graph.
Why is the Velocity-Time graph of a vertical projectile a straight line?
Since the acceleration is constant, the velocity changes at a constant rate over time (v = v0 - gt). The constant rate of change results in a straight line with a constant negative slope equal to -g.
How do you find the velocity of a vertical projectile at any time t?
The velocity can be found using the first equation of motion: v = v0 - gt (where upward is positive).
Can vertical velocity be negative?
Yes. A positive velocity means the object is moving upward, while a negative velocity means it is moving downward.
What are some real-world examples of vertical projectile motion?
Examples include throwing a tennis ball straight up into the air, dropping a stone down a well, a diver jumping straight up off a diving board, or firing a model rocket vertically.