Upward Motion: Finding Time To Reach A Specific Height

Hey guys! Let's dive into a classic physics problem. We're going to figure out how long it takes for a body thrown upwards to reach a certain height. This kind of problem is super common, and understanding it will help you grasp the basics of kinematics. So, buckle up, because we're about to break it down step by step. We'll be using the example of a body thrown upwards with an initial velocity of $20 \text{ m/s}$, and we'll calculate the time it takes to reach a height of $7.2 \text{ m}$. The cool thing is, this problem has two possible answers, which we'll explore. Ready to get started? Let's go!

Understanding the Problem: Vertical Motion

Okay, first things first. What's happening here? We have an object (let's say a ball) that's thrown straight up. Gravity is the only force acting on it (we're ignoring air resistance for simplicity). This means the ball will slow down as it goes up, reach a peak, and then speed up as it comes back down. This type of motion is called vertical motion or projectile motion in one dimension. The key concepts here are initial velocity, acceleration due to gravity, and displacement (the height the ball reaches). This problem is all about understanding how these concepts relate to each other to find time.

When the ball is thrown upwards, it slows down because gravity is pulling it downwards. At the highest point of its trajectory, the ball's velocity momentarily becomes zero before it starts to fall back down. As the ball falls, it accelerates downwards due to gravity. Understanding these principles is crucial for solving the problem correctly. Another important thing to keep in mind is that the acceleration due to gravity ($g$) is constant and acts downwards. We'll typically use $g = 9.8 \text{ m/s}^2$ (or $10 \text{ m/s}^2$ for easier calculations, depending on the problem). This constant acceleration changes the ball's velocity over time. So, how do we go about actually calculating the time it takes to reach a specific height? We use kinematic equations, which relate displacement, initial velocity, time, and acceleration. Let's move on to how we set up the problem.

Setting Up the Problem: Key Variables and Equations

Alright, let's get to the meat of the problem. We're given the initial velocity ($v_0$) of the body, which is $20 \text{ m/s}$. We're also given the height ($h$) we want to find the time for, which is $7.2 \text{ m}$. The acceleration due to gravity ($g$) is $9.8 \text{ m/s}^2$. Since the body is moving upwards, we'll consider upward direction as positive and downward direction as negative. Therefore, the acceleration due to gravity will be $-9.8 \text{ m/s}^2$ (or $-10 \text{ m/s}^2$). Our goal is to find the time ($t$) it takes to reach the height $h$. To do this, we will use one of the standard kinematic equations of motion. This equation directly relates displacement, initial velocity, acceleration, and time:

$h = v_0t + \frac{1}{2}at^2$

Where:

• $h$ is the displacement (height), which is $7.2 \text{ m}$.
• $v_0$ is the initial velocity, which is $20 \text{ m/s}$.
• $a$ is the acceleration due to gravity, which is $-9.8 \text{ m/s}^2$.
• $t$ is the time, which is what we're trying to find.

Plugging in the values, we get:

$7.2 = 20t - \frac{1}{2} \cdot 9.8t^2$

This simplifies to a quadratic equation, which we'll solve next. This equation will give us two possible times. That's because the body passes the height of $7.2 \text{ m}$ twice: once on its way up and once on its way down. This is why we have two answers, $0.4 \text{ s}$ and $3.6 \text{ s}$. Let's see how we solve for the time. We're almost there!

Solving for Time: Quadratic Equation and Solutions

Now, let's solve the quadratic equation derived from the kinematic equation. From the previous step, we have:

$7.2 = 20t - 4.9t^2$

Let's rearrange this equation to standard quadratic form ($at^2 + bt + c = 0$):

$4.9t^2 - 20t + 7.2 = 0$

To solve for $t$, we can use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = 4.9$
• $b = -20$
• $c = 7.2$

Plugging in the values:

$t = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 4.9 \cdot 7.2}}{2 \cdot 4.9}$

$t = \frac{20 \pm \sqrt{400 - 141.12}}{9.8}$

$t = \frac{20 \pm \sqrt{258.88}}{9.8}$

$t = \frac{20 \pm 16.09}{9.8}$

This gives us two possible solutions for $t$:

• $t_1 = \frac{20 + 16.09}{9.8} \approx 3.68 \text{ s}$
• $t_2 = \frac{20 - 16.09}{9.8} \approx 0.399 \text{ s}$

So, we get approximately $0.4 \text{ s}$ and $3.6 \text{ s}$. The first time, $0.4 \text{ s}$, is when the body reaches the height on its way up. The second time, $3.6 \text{ s}$, is when the body reaches the height on its way down. And there you have it! We've solved for the time it takes for the body to reach the specified height. This is a great example of how to apply the kinematic equations to solve real-world physics problems. Let's move on to the conclusion and wrap things up.

Conclusion: The Two Times and Their Significance

Alright, guys, we've made it to the end! We've successfully calculated the time it takes for a body thrown upwards with an initial velocity of $20 \text{ m/s}$ to reach a height of $7.2 \text{ m}$. We found two solutions: approximately $0.4 \text{ s}$ and $3.6 \text{ s}$. These two answers reflect the nature of the upward and downward motion of the body. The first time ($0.4 \text{ s}$) represents the time it takes for the body to reach the height on its way up. The second time ($3.6 \text{ s}$) represents the time it takes for the body to reach the same height again on its way down. This dual solution is a key characteristic of projectile motion. It highlights that the object passes a specific height twice during its flight (unless the height is at the peak).

Understanding these concepts is crucial for any physics student. The ability to correctly interpret the two solutions is a sign that you understand the physics behind the problem. This knowledge can then be extended to more complex situations, such as varying initial velocities and taking air resistance into account. The process we followed – setting up the problem, identifying the key variables, choosing the appropriate kinematic equation, and solving the resulting quadratic equation – is a standard method for solving many physics problems. Keep practicing, and you'll become a pro in no time. Well done, everyone! If you have any questions, feel free to ask. See you in the next physics adventure!