Projectile Motion Problem: Height, Time, And Velocity
Hey guys! Let's dive into a classic physics problem involving projectile motion. We've got a scenario where an object—think of it like an arrow—is launched straight up into the air. We're given its initial speed, and we need to figure out a few things: how high it goes, how long it takes to get there, and where it is (and how fast it's moving) after a certain amount of time. Sounds fun, right? This is a very practical application of physics principles, so understanding this will really help you grasp the concepts of motion and gravity. So, let’s get started and break down this problem step by step!
Understanding the Problem
In this projectile motion problem, the initial upward velocity of our projectile is 9.75 m/s. The big questions we're tackling are: First, what's the maximum height this projectile will reach? This is crucial for understanding the range of its motion. Second, how much time will it take for the projectile to reach its highest point? This helps us understand the duration of the ascent phase. Finally, what will be the projectile's position and velocity after 6 seconds? This question explores the projectile’s motion over time, including its descent if applicable. The key here is recognizing that gravity is the main force acting on the projectile, causing it to decelerate as it moves upward and accelerate as it falls downward. Understanding this dynamic is fundamental to solving the problem accurately.
To solve this, we'll use some fundamental physics equations that describe motion under constant acceleration. Remember, gravity provides a constant downward acceleration (approximately 9.8 m/s²), which we'll denote as 'g'. We'll also need to keep in mind that at the projectile's maximum height, its instantaneous vertical velocity will be zero. This is a critical point that allows us to calculate both the maximum height and the time it takes to reach it. After we've covered those basics, we'll use the same equations to determine the projectile's position and velocity at the 6-second mark. This will give us a complete picture of its journey from launch to a point well into its flight. Let's break down the calculations next!
Calculating Maximum Height
Alright, let's figure out the maximum height our projectile reaches. This involves using a kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The most suitable equation for this scenario is: v_f² = v_i² + 2ad, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the displacement (which will be the maximum height in our case). Remember, guys, at the maximum height, the final velocity (v_f) will be 0 m/s because the projectile momentarily stops before it starts falling back down.
We know that the initial velocity (v_i) is 9.75 m/s, and the acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s² (negative because it acts downwards, opposite to the initial upward motion). Now we can plug these values into our equation: 0 = (9.75 m/s)² + 2(-9.8 m/s²)d. Solving for d will give us the maximum height. Let’s do the math! First, we calculate (9.75 m/s)² which equals 95.0625 m²/s². Our equation now looks like this: 0 = 95.0625 m²/s² - 19.6 m/s² * d. Next, we isolate d by moving the term with d to the left side of the equation: 19.6 m/s² * d = 95.0625 m²/s². Finally, we divide both sides by 19.6 m/s² to solve for d: d = 95.0625 m²/s² / 19.6 m/s². This gives us a maximum height, d, of approximately 4.85 meters. So, our projectile reaches a peak height of about 4.85 meters. Pretty cool, huh? Now, let’s figure out how long it takes to get there!
Calculating Time to Reach Maximum Height
Now, let's figure out how long it takes for the projectile to reach that maximum height. For this, we'll use another kinematic equation, one that directly relates initial velocity, final velocity, acceleration, and time. The equation we're going to use is: v_f = v_i + at, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time. Again, the final velocity (v_f) at the maximum height is 0 m/s.
We already know the initial velocity (v_i) is 9.75 m/s, and the acceleration (a) is -9.8 m/s² (due to gravity). We can plug these values into our equation: 0 m/s = 9.75 m/s + (-9.8 m/s²) * t. Now we need to solve for t. Let’s rearrange the equation to isolate t. First, subtract 9.75 m/s from both sides: -9.75 m/s = -9.8 m/s² * t. Then, divide both sides by -9.8 m/s² to solve for t: t = (-9.75 m/s) / (-9.8 m/s²). This gives us a time, t, of approximately 0.995 seconds. So, it takes about 0.995 seconds, or roughly 1 second, for the projectile to reach its maximum height. Not too shabby! Next up, we'll figure out its position and velocity after 6 seconds.
Determining Position and Velocity After 6 Seconds
Okay, guys, this is where it gets a little more interesting! We need to find the projectile's position and velocity after 6 seconds. Remember, the projectile went up and came back down, and 6 seconds is quite a bit of time compared to the roughly 1 second it took to reach its peak. This means our projectile will be on its way down.
First, let's find the position. We can use the kinematic equation: d = v_it + (1/2)at², where d is the displacement, v_i is the initial velocity, t is the time (6 seconds in this case), and a is the acceleration due to gravity (-9.8 m/s²). Plugging in the values, we get: d = (9.75 m/s)(6 s) + (1/2)(-9.8 m/s²)(6 s)². Let's calculate this step by step. First, (9.75 m/s)(6 s) equals 58.5 meters. Then, (6 s)² is 36 s², and (1/2)(-9.8 m/s²)(36 s²) equals -176.4 meters. So, our equation now looks like this: d = 58.5 m - 176.4 m. This gives us a displacement, d, of -117.9 meters. The negative sign indicates that the projectile is 117.9 meters below its starting point after 6 seconds.
Now, let's find the velocity after 6 seconds. We can use the equation: v_f = v_i + at. We know v_i is 9.75 m/s, a is -9.8 m/s², and t is 6 s. Plugging these in, we get: v_f = 9.75 m/s + (-9.8 m/s²)(6 s). Calculating this, (-9.8 m/s²)(6 s) equals -58.8 m/s. So, v_f = 9.75 m/s - 58.8 m/s, which gives us a final velocity of -49.05 m/s. The negative sign here indicates that the velocity is directed downwards. Therefore, after 6 seconds, the projectile is approximately 117.9 meters below its starting point and is moving downwards at a speed of 49.05 m/s. See how it all comes together? Now, let's wrap up with a final summary.
Final Summary
Alright, let's recap what we've discovered in this projectile motion problem! We started with a projectile launched upwards at an initial velocity of 9.75 m/s, and we wanted to know its maximum height, the time it took to reach that height, and its position and velocity after 6 seconds. Using our kinematic equations, we found that the maximum height the projectile reaches is approximately 4.85 meters. It takes about 0.995 seconds (or roughly 1 second) for the projectile to reach this peak.
After 6 seconds, things get interesting! The projectile has already reached its peak and is falling back down. We calculated that after 6 seconds, the projectile is approximately 117.9 meters below its starting point. This negative displacement tells us it has fallen well past its initial launch position. The velocity at this time is -49.05 m/s, with the negative sign indicating that it's moving downwards. So, at 6 seconds, the projectile is quite far down and moving pretty fast in the downward direction. This problem illustrates beautifully how gravity affects the motion of objects in flight. We've used key physics principles and equations to break down each aspect of the projectile's journey. I hope you guys found this explanation helpful and that it gives you a solid understanding of projectile motion! Remember, physics is all about understanding the world around us, one problem at a time!