Introduction
Hey guys! Today, we're diving into a classic physics problem involving projectile motion. We've got an object launched from a table, and we're going to break down how to calculate its maximum height, time of flight, velocity, and position at different points in its trajectory. This is a fundamental concept in physics, and mastering it will help you understand a wide range of real-world scenarios, from sports to engineering. So, let's get started and make physics fun!
This article provides a detailed solution to a projectile motion problem, covering essential concepts and calculations. We'll walk through each step in a friendly, conversational manner, ensuring you grasp the underlying principles. Whether you're a student tackling homework or just curious about physics, this guide is here to help you understand the mechanics of projectile motion.
Problem Statement
Alright, let's lay out the problem. Imagine we have an object that's launched from a table that's 5 meters high. It's launched with an initial velocity of 10 meters per second. Our mission, should we choose to accept it (and we do!), is to figure out a few key things about its journey:
- (a) The maximum height, H: How high does this object go above the ground?
- (b) The time of flight: How long is the object in the air before it hits the ground?
- (c) Its velocity and height when t = 2 s: What's the object's speed and altitude exactly 2 seconds after launch?
- (d) Its velocity when it hits the ground: How fast is the object moving when it finally lands?
This is a classic projectile motion problem that involves understanding the effects of gravity on an object's movement. We will break down the problem into manageable steps, using kinematic equations to find the solutions. Don't worry if it sounds complicated right now; we'll take it slow and make sure everything clicks.
(a) Maximum Height Calculation
To figure out the maximum height the object reaches, we need to think about what happens at the very top of its flight. At that instant, the object's vertical velocity is momentarily zero. We can use this fact, along with some kinematic equations, to find the maximum height.
The key here is to break the initial velocity into its vertical and horizontal components. Since we're only interested in the vertical motion for this part, let's focus on the vertical component. Let's call the initial vertical velocity v₀y. We'll need to know the launch angle to calculate this, but for now, let's assume we have that angle (θ). Then, v₀y = 10 m/s * sin(θ). If the launch angle is not provided, we will assume it is launched vertically upwards (θ = 90 degrees), making sin(θ) = 1.
Now, we can use the following kinematic equation:
v² = v₀² + 2 * a * Δy
Where:
- v is the final vertical velocity (0 m/s at maximum height)
- v₀ is the initial vertical velocity (v₀y)
- a is the acceleration due to gravity (-9.8 m/s²)
- Δy is the vertical displacement (the height we want to find)
Plugging in the values, we get:
0 = (10 m/s * sin(θ))² + 2 * (-9.8 m/s²) * Δy
Solving for Δy, which represents the height above the launch point, we have:
Δy = (10 m/s * sin(θ))² / (2 * 9.8 m/s²)
If we assume the launch is straight up, this simplifies to:
Δy = (10 m/s)² / (2 * 9.8 m/s²) ≈ 5.1 meters
But remember, this is the height above the table. To find the maximum height above the ground, we need to add the table's height:
H = 5.1 meters + 5 meters = 10.1 meters
So, the maximum height the object reaches above the ground is approximately 10.1 meters. This calculation highlights the importance of understanding the vertical motion component and using the appropriate kinematic equations to solve for unknowns.
(b) Time of Flight Calculation
Next up, let's calculate the time of flight. This is the total time the object spends in the air, from the moment it's launched until it hits the ground. To figure this out, we need to consider the entire vertical journey, both upwards and downwards.
We can use another kinematic equation that relates displacement, initial velocity, time, and acceleration:
Δy = v₀y * t + 0.5 * a * t²
Where:
- Δy is the total vertical displacement (in this case, -5 meters, since the object ends up 5 meters below its starting point)
- v₀y is the initial vertical velocity (10 m/s * sin(θ))
- a is the acceleration due to gravity (-9.8 m/s²)
- t is the time of flight (what we want to find)
Plugging in the values, we get:
-5 meters = (10 m/s * sin(θ)) * t + 0.5 * (-9.8 m/s²) * t²
Assuming a vertical launch, this simplifies to:
-5 = 10 * t - 4.9 * t²
This is a quadratic equation, and we need to solve it for t. Rearranging the equation, we get:
- 9 * t² - 10 * t - 5 = 0
We can use the quadratic formula to solve for t:
t = [-b ± √(b² - 4ac)] / (2a)
Where:
- a = 4.9
- b = -10
- c = -5
Plugging in these values, we get two possible solutions for t. We'll take the positive solution since time cannot be negative.
t = [10 ± √((-10)² - 4 * 4.9 * -5)] / (2 * 4.9)
t ≈ [10 ± √(100 + 98)] / 9.8
t ≈ [10 ± √198] / 9.8
t ≈ (10 ± 14.07) / 9.8
We have two solutions: t ≈ 2.46 seconds and t ≈ -0.42 seconds. Since time cannot be negative, we discard the negative solution. Thus:
t ≈ 2.46 seconds
Therefore, the time of flight is approximately 2.46 seconds. This calculation demonstrates how quadratic equations come into play when dealing with projectile motion over varying heights.
(c) Velocity and Height at t = 2 s Calculation
Now, let's figure out the velocity and height of the object at t = 2 seconds. This requires us to look at both the vertical and horizontal components of the motion.
Vertical Motion
First, let's find the vertical velocity at t = 2 s. We can use the following kinematic equation:
v_y = v₀y + a * t
Where:
- v_y is the vertical velocity at time t
- v₀y is the initial vertical velocity (10 m/s * sin(θ))
- a is the acceleration due to gravity (-9.8 m/s²)
- t = 2 s
Assuming a vertical launch:
v_y = 10 m/s + (-9.8 m/s²) * 2 s
v_y = 10 m/s - 19.8 m/s
v_y ≈ -9.8 m/s
The negative sign indicates that the object is moving downwards at this point.
Next, let's find the height at t = 2 s using the equation:
y = v₀y * t + 0.5 * a * t² + y₀
Where:
- y is the height at time t
- v₀y is the initial vertical velocity (10 m/s)
- a is the acceleration due to gravity (-9.8 m/s²)
- t = 2 s
- y₀ is the initial height (5 meters)
y = (10 m/s * 2 s) + 0.5 * (-9.8 m/s²) * (2 s)² + 5 meters
y = 20 meters - 19.6 meters + 5 meters
y ≈ 5.4 meters
Horizontal Motion
Since there is no horizontal acceleration (we're neglecting air resistance), the horizontal velocity remains constant throughout the motion. If the launch angle is θ, the initial horizontal velocity v₀x is given by:
v₀x = 10 m/s * cos(θ)
If the launch is vertical, the horizontal velocity is 0 m/s.
Resultant Velocity
The resultant velocity at t = 2 s is the vector sum of the vertical and horizontal velocities. If we assume a vertical launch, the horizontal velocity is zero, and the resultant velocity is just the vertical velocity, which is approximately -9.8 m/s (downwards).
If there's a horizontal component, the resultant velocity (v) can be found using the Pythagorean theorem:
v = √(v_x² + v_y²)
The direction of the velocity can be found using trigonometry:
Angle = tan⁻¹(v_y / v_x)
In summary, at t = 2 seconds, the object is approximately 5.4 meters above the ground and moving downwards at about 9.8 m/s (if launched vertically). This section highlights the importance of considering both vertical and horizontal components of motion to fully describe the object's state at a given time.
(d) Velocity When It Hits the Ground Calculation
Finally, let's calculate the velocity of the object when it hits the ground. We're interested in the magnitude and direction of the velocity at the moment of impact.
Vertical Velocity
We can use the following kinematic equation to find the final vertical velocity (v_fy) just before impact:
v_fy² = v₀y² + 2 * a * Δy
Where:
- v_fy is the final vertical velocity
- v₀y is the initial vertical velocity (10 m/s * sin(θ))
- a is the acceleration due to gravity (-9.8 m/s²)
- Δy is the total vertical displacement (-5 meters, as the object lands 5 meters below its starting point)
Assuming a vertical launch:
v_fy² = (10 m/s)² + 2 * (-9.8 m/s²) * (-5 meters)
v_fy² = 100 m²/s² + 98 m²/s²
v_fy² = 198 m²/s²
v_fy = ±√198 m²/s²
v_fy ≈ ±14.07 m/s
Since the object is moving downwards when it hits the ground, we take the negative root:
v_fy ≈ -14.07 m/s
Horizontal Velocity
As before, the horizontal velocity remains constant throughout the motion. If the initial horizontal velocity (v₀x) is 10 m/s * cos(θ), it remains the same until impact. For a vertical launch, v₀x is 0 m/s.
Resultant Velocity
The resultant velocity (v_f) when the object hits the ground is the vector sum of the final vertical and horizontal velocities. If we assume a vertical launch, the resultant velocity is just the final vertical velocity, which is approximately -14.07 m/s (downwards).
If there's a horizontal component, the resultant velocity can be found using the Pythagorean theorem:
v_f = √(v_fx² + v_fy²)
And the angle of impact can be found using:
Angle = tan⁻¹(v_fy / v_fx)
Therefore, the object hits the ground with a velocity of approximately 14.07 m/s downwards (for a vertical launch). This final calculation ties together all the concepts we've discussed, giving us a complete picture of the object's motion from launch to impact.
Conclusion
So, there you have it! We've successfully broken down a projectile motion problem and calculated the maximum height, time of flight, velocity at a specific time, and impact velocity. This problem illustrates the power of kinematic equations and the importance of understanding the vertical and horizontal components of motion. Projectile motion might seem tricky at first, but with a clear understanding of the principles and a step-by-step approach, you can tackle these problems like a pro.
Remember, physics is all about understanding the world around us, and projectile motion is a key concept in many real-world scenarios. Keep practicing, and you'll be mastering these concepts in no time! If you found this guide helpful, give it a share and let's keep the physics learning going!