Vertical Throw Simulation – Explore Upward Motion Under Gravity

Vertical upward motion is the movement of an object thrown upward with an initial velocity V₀ in the Earth's gravitational field. In other words, it is the motion of an object along the Y-axis in the XOY plane under the influence of gravity, with velocity V, where the initial velocity V₀ forms a 90° angle with the X-axis.

The velocity of the object decreases over time t by g*t due to the gravitational force pulling the object downward (free fall). After some time, the object reaches its maximum height H_MaxV, and then begins to fall freely downward. On the Earth's surface, the gravitational acceleration is approximately g = 9.81 m/s².

Determining Maximum Height

Let's assume we have a vertical upward throw, and the object is given an initial velocity v0. To determine the maximum height during vertical motion, we can use either the equations of uniformly decelerated motion, or the principle of conservation of mechanical energy.

Conservation of Energy

The total mechanical energy of a body with mass m at height H is the sum of its potential and kinetic energy:

E = Ep + Ek

Where:

• Potential energy at height H:
  

  Ep = m · g · H

  (g is the gravitational acceleration, approximately 9.81 m/s²).
• Kinetic energy of the object moving at speed v:
  

  Ek = ½ · m · v²

At the top of the trajectory (maximum height), the vertical speed becomes zero (v = 0), so the kinetic energy is also zero. At the moment of release (height H = 0), all the energy is in the form of kinetic energy:

Etotal (initial) = ½ · m · v0²

As the object ascends, its kinetic energy decreases and is converted into potential energy. This transformation can be observed in the energy diagrams of the simulation (see figure 2). At the highest point (Hmax), all kinetic energy has been transformed into height:

½ · m · v0² = m · g · Hmax

From this equation, we can directly derive the formula for the maximum height:

Hmax = v0² / (2 · g)

Using Kinematic Equations

Alternatively, we can use the kinematic equation for uniformly decelerated motion (in the vertical direction). Ignoring air resistance, the vertical acceleration is −g. The initial vertical velocity is v0, and at the top of the motion the velocity becomes zero.

v² = v0² − 2 · g · ΔH

At the top of the trajectory, v = 0, so:

0 = v0² − 2 · g · Hmax

and therefore:

Hmax = v0² / (2 · g)

Conclusion

Regardless of whether we use the conservation of energy or kinematic equations, the final result for the maximum height in vertical motion is the same:

Hmax = v0²/(2 · g)

This expression shows that the square of the initial speed v0 significantly influences the maximum height an object can reach, while a higher gravitational acceleration g reduces the maximum height.

Falling of the Body After Reaching the Maximum Height Hmax

When an object is thrown vertically upward, it slows down under the influence of gravity until it reaches zero velocity at the top of its trajectory. This height is denoted as Hmax. At that moment, the object momentarily comes to rest, and then begins to fall back down toward the ground. The further motion can be considered a free fall from the height Hmax, since the vertical velocity is 0 and only gravity acts on the object.

In other words, after the velocity drops to v = 0 at height Hmax, the object immediately starts accelerating downward toward the ground, following the same principle as in free fall from a given height H0. At that moment, the total mechanical energy is distributed as:

• Potential energy: Ep = m · g · Hmax, since the object is at the highest point of the trajectory.
• Kinetic energy: Ek = 0, because the velocity at that moment is v = 0.

From the moment the object starts falling from Hmax, all formulas and conclusions we already mentioned for free fall from a given height apply. This means the object will cover the same distance downward and reach an impact velocity of v = √[2 · g · Hmax], just as if it had been dropped directly from height Hmax.

Object's Speed During a Fall. Kinetic and Potential Energy During Free Fall

When an object is released from a height H0 with no initial velocity, it accelerates due to gravity. At any given moment, part of the total mechanical energy of the object remains as potential energy, while the rest is converted into kinetic energy. The following data explains how speed changes during free fall and how energy is distributed.

1. Potential Energy at Height h

At a moment when the object is at height h above the ground (where 0 ≤ h ≤ H0), its potential energy is:

Ep(h) = m · g · h

As the object falls, h decreases, and the potential energy decreases proportionally with the height.

2. Kinetic Energy and Speed at Height h

Since the total mechanical energy of the object is conserved (neglecting air resistance), the following holds:

Etotal = Ep(h) + Ek(h) = const = m · g · H0

The kinetic energy of the object when it is at height h is:

Ek(h) = m · g · (H0 − h)

From the kinetic energy, we can express the speed v(h):

½ · m · v(h)² = m · g · (H0 − h)  
⇒  
v(h) = √2 · g · (H0 − h)

Therefore, the smaller the h (the closer the object is to the ground), the greater the speed. At the ground level (h = 0), the speed reaches:

vcritical = √2 · g · H0

3. Motion Over Time

In classical mechanics, the speed during free fall depends on the time t elapsed since the object was released:

v(t) = g · t

The height of the object at that moment is:

h(t) = H0 − ½ · g · t²

When h(t) = 0 (the object reaches the ground), the fall time tfall is obtained from:

0 = H0 − ½ · g · tfall²  
⇒  
tfall = √2 · H0 / g

At this time, the speed of the object is:

vfall = g · tfall = √2 · g · H0

4. Energy Over Time

- At the initial moment (t = 0), all energy is in the form of potential energy: Ep = m · g · H0, Ek = 0.
- As the object falls, Ep decreases linearly with height, and Ek increases so that the total remains constant.
- At any moment t and height h(t), we have:

Ep(t) = m · g · h(t)  
Ek(t) = ½ · m · [v(t)]² = ½ · m · (g · t)²  
Etotal = m · g · H0   (constant)

5. Conclusion

Thus, during the free fall of an object: - Speed increases continuously until the object reaches the ground, following the formula v = √2 · g · (H0 − h) or v(t) = g · t.
- Potential energy is converted into kinetic energy, but the total energy remains constant (Ep + Ek = const) until impact.
- At the ground level (h = 0), the object reaches its maximum kinetic energy ½ · m · [√(2 · g · H0)]² = m · g · H0.

Vertical Projectile Motion – Speed and Height

The following formulas describe vertical projectile motion:

Vertical velocity component: v_0Y = v₀ · sin(α) − g · t

Vertical displacement over time: y = v₀ · t − (g · t²) / 2

Time to reach the highest point: t = v₀ / g

Maximum height reached: H_Max = v₀² / (2 · g)