Projectile Motion: Height Of Vertical Projectile Explained

Let's dive into a classic physics problem involving projectile motion! We're going to break down what happens when you launch two projectiles at the same time – one straight up and the other horizontally. The big question we're tackling is: what's the height of the projectile shot upwards when the horizontal one finally hits the ground, assuming air resistance isn't a factor?

Understanding the Scenario

Okay, guys, picture this. You've got two projectiles. Projectile A is launched vertically, battling gravity as it climbs skyward. Projectile B is launched horizontally, speeding forward while also being pulled down by gravity. We're ignoring air resistance to keep things simple, which means we're only dealing with the constant force of gravity acting downwards. This is a common simplification in introductory physics to make the calculations manageable and focus on the core concepts.

The key here is realizing that time is the great connector. Both projectiles are launched simultaneously, and we're interested in the vertical projectile's height at the instant the horizontal projectile lands. This means we need to figure out how long it takes the horizontal projectile to hit the ground, and then use that time to determine the vertical projectile's position.

Initial Conditions: Let's define some variables to make our lives easier:

• v₀ᵥ: Initial vertical velocity of projectile A (the one shot upwards).
• v₀ₕ: Initial horizontal velocity of projectile B (the one shot horizontally).
• h: Initial height from which projectile B is launched (this is crucial for determining its time of flight).
• g: Acceleration due to gravity (approximately 9.8 m/s²).

Understanding these initial conditions and how they influence each projectile's motion is crucial for solving the problem. We're essentially dissecting the motion into its vertical and horizontal components, analyzing each separately, and then connecting them through the common variable of time. Remember, gravity only acts vertically, so the horizontal velocity of projectile B remains constant throughout its flight (since we're neglecting air resistance).

Analyzing the Horizontal Projectile

The horizontal motion of projectile B is super straightforward because there's no horizontal acceleration (again, no air resistance!). It travels at a constant horizontal velocity, v₀ₕ. To find the time it takes to hit the ground, we only need to consider its vertical motion. The horizontal distance it covers is irrelevant to the time it takes to fall. The horizontal distance would only matter if we were trying to figure out how far away it lands from the launch point.

The vertical motion of projectile B is uniformly accelerated due to gravity. We can use the following kinematic equation to find the time, t, it takes to hit the ground:

h = v₀ᵥt + (1/2)gt²

Wait a minute! You might be thinking, "But projectile B is launched horizontally, so its initial vertical velocity is zero!" And you'd be absolutely right! That simplifies our equation to:

h = (1/2)gt²

Now, we can solve for t:

t = √(2h/g)

This equation tells us the time it takes for the horizontal projectile to hit the ground, depending only on the initial height and the acceleration due to gravity. Notice that the initial horizontal velocity, v₀ₕ, doesn't appear in this equation. That's because it only affects how far the projectile travels horizontally, not how long it takes to fall.

Key Takeaway: The time it takes for the horizontal projectile to hit the ground only depends on the height from which it was launched and the acceleration due to gravity. The initial horizontal velocity is irrelevant for this calculation.

Analyzing the Vertical Projectile

Now, let's switch gears and focus on projectile A, the one launched straight up. Its motion is also uniformly accelerated due to gravity, but this time, gravity is working against its initial upward velocity. We want to find its height at the same time we calculated for the horizontal projectile to hit the ground, t = √(2h/g).

We can use another kinematic equation to find the vertical displacement, y, of projectile A at time t:

y = v₀ᵥt - (1/2)gt²

Notice the minus sign in front of the (1/2)gt² term. This is because gravity is acting downwards, opposing the initial upward velocity. Now, we simply plug in the value of t we found earlier:

y = v₀ᵥ√(2h/g) - (1/2)g(2h/g)

Simplifying the equation, we get:

y = v₀ᵥ√(2h/g) - h

This equation gives us the height, y, of the vertical projectile at the instant the horizontal projectile hits the ground. It depends on the initial vertical velocity (v₀ᵥ), the initial height of the horizontal projectile (h), and the acceleration due to gravity (g).

Important Considerations:

• If v₀ᵥ√(2h/g) < h, then y will be negative, meaning the vertical projectile has already fallen back to the ground (or even below the initial launch height) by the time the horizontal projectile lands.
• If v₀ᵥ√(2h/g) = h, then y will be zero, meaning the vertical projectile is at ground level when the horizontal projectile lands.
• If v₀ᵥ√(2h/g) > h, then y will be positive, meaning the vertical projectile is still in the air when the horizontal projectile lands.

Putting It All Together

So, to recap, guys, here's the breakdown:

1. Horizontal Projectile: We calculated the time it takes for the horizontal projectile to hit the ground using the equation t = √(2h/g), where h is the initial height and g is the acceleration due to gravity.
2. Vertical Projectile: We used this time to find the height of the vertical projectile using the equation y = v₀ᵥ√(2h/g) - h, where v₀ᵥ is the initial vertical velocity.

Final Answer: The height of the vertical projectile at the instant the horizontal projectile hits the ground is given by the equation y = v₀ᵥ√(2h/g) - h.

Key Concepts Revisited:

• Independence of Motion: The horizontal and vertical motions of a projectile are independent of each other. This means we can analyze them separately and then combine the results using time as the common variable.
• Uniform Acceleration: The vertical motion of both projectiles is uniformly accelerated due to gravity. This allows us to use kinematic equations to relate displacement, velocity, time, and acceleration.
• Constant Horizontal Velocity: The horizontal motion of the horizontal projectile is at a constant velocity (assuming no air resistance). This simplifies the analysis of its horizontal motion.

Example Scenario

Let's plug in some numbers to make this even clearer. Suppose the horizontal projectile is launched from a height of 20 meters (h = 20 m), and the vertical projectile is launched with an initial velocity of 15 m/s (v₀ᵥ = 15 m/s). What is the height of the vertical projectile when the horizontal projectile hits the ground?

First, we calculate the time it takes for the horizontal projectile to hit the ground:

t = √(2h/g) = √(2 * 20 m / 9.8 m/s²) ≈ 2.02 seconds

Now, we plug this time into the equation for the height of the vertical projectile:

y = v₀ᵥ√(2h/g) - h = 15 m/s * √(2 * 20 m / 9.8 m/s²) - 20 m ≈ 15 m/s * 2.02 s - 20 m ≈ 10.3 meters

So, in this example, the vertical projectile is approximately 10.3 meters above the ground when the horizontal projectile hits the ground.

Common Mistakes to Avoid

• Forgetting the Initial Height: Make sure you account for the initial height of the horizontal projectile when calculating its time of flight.
• Mixing Up Velocities: Be careful to distinguish between initial vertical and horizontal velocities. Remember that the initial horizontal velocity doesn't affect the time it takes to fall, and the initial vertical velocity affects the height the projectile reaches.
• Incorrectly Applying Kinematic Equations: Double-check that you're using the correct kinematic equations and that you're plugging in the values correctly.
• Ignoring the Direction of Gravity: Remember that gravity acts downwards, so the acceleration due to gravity is negative when the upward direction is considered positive.

Conclusion

Understanding projectile motion involves breaking down the motion into its horizontal and vertical components and then analyzing each component separately. By using the principles of uniform acceleration and constant velocity, and by carefully applying kinematic equations, we can solve a wide variety of projectile motion problems. Remember to pay attention to the initial conditions, and always double-check your work to avoid common mistakes.

So there you have it, folks! Hopefully, this explanation has cleared up any confusion about this projectile motion problem. Keep practicing, and you'll be a projectile motion pro in no time!