Projectiles Launched From Moving Vehicles: Videos & Practice Problems

Projectile motion can occur from moving vehicles, where the projectile inherits the vehicle's velocity. For example, if a package is released from a plane traveling at 300 m/s, it also moves at 300 m/s horizontally. When launching from a downward-moving balloon at 3 m/s with an outward velocity of 4 m/s, vector addition gives a resultant velocity of 5 m/s. The equation for the projectile's velocity is $v^{\mathrm{projectile}}=v^{\mathrm{launch}}+v^{\mathrm{vehicle}}$. This approach applies to all projectile motion problems.

Releasing or Launching Projectiles From Moving Vehicles

In projectile motion problems involving moving vehicles, it is essential to differentiate between the projectile and the vehicle from which it is launched. The projectile is the object being released, while the vehicle is the one in motion. The velocity of the vehicle is denoted as \( v_{\text{vehicle}} \). When a projectile is released from a moving vehicle, it inherits the vehicle's velocity at the moment of release, which means that the initial velocity of the projectile, \( v_{\text{projectile}} \), is a combination of the vehicle's velocity and the launch velocity of the projectile.

For instance, if a package is released from an airplane traveling at 300 meters per second, the package will also have an initial horizontal velocity of 300 meters per second. Once released, the projectile follows a parabolic path influenced solely by gravity, behaving like any other horizontally launched projectile. In another scenario, if a balloon is descending at 3 meters per second while launching an object outward at 4 meters per second, the resultant velocity of the projectile can be determined by vector addition. The downward velocity of the balloon and the outward launch velocity combine to create a two-dimensional velocity vector, which can be calculated using the Pythagorean theorem:

$$ v_{\text{projectile}} = \sqrt{(v_{\text{launch}}^2 + v_{\text{vehicle}}^2)} $$

In this case, the projectile's velocity would be \( \sqrt{(4^2 + 3^2)} = 5 \) meters per second.

When launching a projectile straight upwards from a moving vehicle, such as a car moving at 30 meters per second while launching an object at 40 meters per second, the same principle applies. The resultant velocity of the projectile is again found using the Pythagorean theorem, resulting in a velocity of 50 meters per second, as it combines both the upward and horizontal components.

In summary, the velocity of the projectile can be expressed as:

$$ v_{\text{projectile}} = v_{\text{launch}} + v_{\text{vehicle}} $$

In cases where the projectile is simply dropped, the launch velocity is zero, and the projectile moves with the same velocity as the vehicle.

To solve for the maximum height of a projectile launched from a moving vehicle, one can follow the same steps as in standard projectile motion problems. For example, if a cart moves at 60 meters per second and launches a missile upwards at 80 meters per second, the initial vertical velocity is 80 meters per second. The maximum height can be calculated using the kinematic equation:

$$ v_{by}^2 = v_{ay}^2 + 2a_y \Delta y $$

Where \( v_{by} \) is the final vertical velocity (0 at the peak), \( v_{ay} \) is the initial vertical velocity (80 m/s), \( a_y \) is the acceleration due to gravity (-9.8 m/s²), and \( \Delta y \) is the change in height. Solving this gives:

$$ 0 = (80)^2 + 2(-9.8)(\Delta y) $$

From which we find that the maximum height \( \Delta y \) is approximately 326.5 meters. This approach illustrates that once the initial velocity of the projectile is determined, the same kinematic equations can be applied as in traditional projectile motion problems.

A small plane flies horizontally at 20m/s at an altitude of 200m, when it launches a projectile at a speed of 65 m/s at 22.6° below the horizontal. What horizontal distance does the projectile travel before hitting the ground?

• A. 752 m
• B. 344 m
• C. 184 m
• D. 920 m

Answer: (B)

Here’s what students ask on this topic:

How do you calculate the velocity of a projectile launched from a moving vehicle?

To calculate the velocity of a projectile launched from a moving vehicle, you need to use vector addition. The projectile's velocity is the sum of the vehicle's velocity and the launch velocity. Mathematically, this is expressed as:

$v$_projectile = v_launch + v_vehicle

For example, if a vehicle is moving at 30 m/s and a projectile is launched upwards at 40 m/s, the resultant velocity is found using the Pythagorean theorem:

$v$_projectile = √(v_launch² + v_vehicle²) = √(30² + 40²) = 50 m/s

What happens to a projectile when it is released from a moving airplane?

When a projectile is released from a moving airplane, it inherits the horizontal velocity of the airplane. For instance, if the airplane is moving at 300 m/s, the projectile will also move horizontally at 300 m/s upon release. After release, the projectile is subjected only to the influence of gravity, causing it to follow a parabolic trajectory. This behavior is similar to any other horizontal launch problem, where the initial horizontal velocity remains constant, and the vertical motion is influenced by gravity.

How do you solve projectile motion problems involving moving vehicles?

To solve projectile motion problems involving moving vehicles, follow these steps:

1. Determine the initial velocity of the projectile by combining the vehicle's velocity and the launch velocity using vector addition.
2. Decompose the initial velocity into horizontal and vertical components.
3. Use kinematic equations to analyze the motion in the horizontal and vertical directions separately.
4. Apply the same steps and equations used in standard projectile motion problems, such as calculating time of flight, maximum height, and range.

For example, if a cart moving at 60 m/s launches a missile upwards at 80 m/s, the initial velocity of the projectile is 100 m/s (using the Pythagorean theorem). Then, decompose this velocity and solve for the desired quantities.

What is the effect of a downward-moving vehicle on the velocity of a launched projectile?

When a projectile is launched from a downward-moving vehicle, the projectile's velocity is the vector sum of the vehicle's downward velocity and the launch velocity. For example, if a balloon is descending at 3 m/s and a projectile is launched outward at 4 m/s, the resultant velocity is calculated using vector addition:

$v$_projectile = √(v_launch² + v_vehicle²) = √(4² + 3²) = 5 m/s

This resultant velocity is then used to analyze the projectile's motion, considering both horizontal and vertical components.

How do you determine the maximum height of a projectile launched from a moving vehicle?

To determine the maximum height of a projectile launched from a moving vehicle, follow these steps:

1. Calculate the initial velocity of the projectile using vector addition of the vehicle's velocity and the launch velocity.
2. Decompose the initial velocity into horizontal and vertical components.
3. Use the vertical component of the initial velocity to find the maximum height using the kinematic equation:

$v$_f² = v_i² + 2aΔy

where $v$_f = 0 at the maximum height, $a=\; -9.8\; m/s2$, and $v$_i is the vertical component of the initial velocity. Solve for $\mathrm{\Delta y}$ to find the maximum height.