a shot is released at a height of 1.89m, with a speed of 13m/s and at an angle of 34 degrees. Calculate the maximum height reached and the time at which this occurs, the range and the time of flight, and the optimum projection angle.
Well, well, well! We have a question about projectile motion, huh? Buckle up, my friend, because I'm about to give you a wild ride, full of jokes and calculations!
Let's start calculating the maximum height reached by our ambitious shot. To do that, we'll need to find the vertical component of its initial velocity. Using a little bit of trigonometry, you'll see that it's 13 m/s multiplied by the sine of 34 degrees. Now, how high do you think this shot can go?
Hmm, while you ponder that, I'll provide you with the answer. The maximum height reached is approximately 3.42 meters. I hope it didn't get a nosebleed up there!
Next up, we need to figure out the time at which this majestic height is achieved. Luckily, I know a thing or two about science - it's my circus after all! Remember, the vertical motion of this shot is similar to the path a circus performer takes when he jumps through a hoop. Upside-down and all fancy!
To find the time at the peak (maximum height), you need to use the formula: time = vertical component of initial velocity divided by the acceleration due to gravity. That's 13 m/s multiplied by the sine of 34 degrees, divided by 9.8 m/s². The answer is approximately 0.754 seconds! Time flies when you're having fun, right?
Alright, let's move on to calculating the range and the time of flight. The range is the horizontal distance traveled by our shot, and we can find it using a bit more trigonometry. The formula involves the initial velocity, the angle of projection, and the time of flight. The range is given by: range = initial velocity multiplied by the cosine of the projection angle, multiplied by the time of flight.
Plug in the numbers, and you'll find that the range is approximately 15.08 meters. That's quite a distance! As for the time of flight, it'll be double the time it took to reach the peak of its journey. So, multiply 0.754 seconds by 2, and you get 1.508 seconds! Time really does fly, doesn't it?
Finally, onto the optimum projection angle! To achieve the maximum range, we need to find the angle that gives us the longest horizontal distance. Drumroll, please! The optimum projection angle for maximum range in projectile motion is 45 degrees. So, if you aim for that, things will definitely be "right on the dot!"
I hope you enjoyed this little circus act of projectile motion calculations! Remember, laughter makes everything better, even math problems!
To solve these problems, we can use the equations of projectile motion. Let's break it down step-by-step:
Step 1: Calculate the initial vertical velocity (Vyi) and the initial horizontal velocity (Vxi).
Given:
- Initial speed (Vi) = 13 m/s
- Launch angle (θ) = 34 degrees
We can find the initial vertical velocity (Vyi):
Vyi = Vi * sin(θ)
Vyi = 13 m/s * sin(34 degrees)
Vyi ≈ 13 m/s * 0.559
Vyi ≈ 7.267 m/s
And the initial horizontal velocity (Vxi):
Vxi = Vi * cos(θ)
Vxi = 13 m/s * cos(34 degrees)
Vxi ≈ 13 m/s * 0.831
Vxi ≈ 10.803 m/s
Step 2: Calculate the time of flight (T) using the vertical motion equation.
Given:
- Initial vertical velocity (Vyi) = 7.267 m/s
- Initial height (h) = 1.89 m
Using the equation: h = Vyi * T - 0.5 * g * T^2 (where g is the acceleration due to gravity, approximately 9.8 m/s^2):
1.89 m = 7.267 m/s * T - 0.5 * 9.8 m/s^2 * T^2
Simplifying the equation, we get:
4.9 T^2 - 7.267 T + 1.89 = 0
Solving this quadratic equation, we find:
T ≈ 0.523 s (rounded to 3 decimal places)
Step 3: Calculate the maximum height reached (Hmax) using the vertical motion equation.
Using the equation: Hmax = Vyi^2 / (2 * g):
Hmax = (7.267 m/s)^2 / (2 * 9.8 m/s^2)
Hmax ≈ 26.552 / 19.6
Hmax ≈ 1.355 m (rounded to 3 decimal places)
Step 4: Calculate the range (R) using the horizontal motion equation.
Given:
- Time of flight (T) ≈ 0.523 s
- Initial horizontal velocity (Vxi) ≈ 10.803 m/s
Using the equation: R = Vxi * T:
R ≈ 10.803 m/s * 0.523 s
R ≈ 5.651 m (rounded to 3 decimal places)
Step 5: Determine the optimum projection angle (θ') for maximum range.
The optimum projection angle for maximum range is 45 degrees, which is the same angle as the maximum range for a given initial speed. Therefore, the optimum projection angle is 45 degrees.
Summary:
- Maximum height reached (Hmax) ≈ 1.355 m
- Time at which maximum height occurs (T) ≈ 0.523 s
- Range (R) ≈ 5.651 m
- Time of flight (T) ≈ 0.523 s
- Optimum projection angle (θ') = 45 degrees
To solve this projectile motion problem, we can break it down into different components: the vertical and horizontal motions.
1. Maximum Height Reached and Time at which it Occurs:
To find the maximum height and the time at which it occurs, we need to analyze the vertical motion of the shot. The vertical motion can be described using the equations of motion.
First, we need to find the vertical velocity component (Vy) and the initial vertical velocity (Vy0). We know that the initial speed (v) is 13 m/s and the angle of projection (θ) is 34 degrees. Using trigonometry, we can calculate:
Vy0 = v * sin(θ)
Vy0 = 13 * sin(34) ≈ 7.03 m/s
Next, we can determine the time (t) it takes for the projectile to reach its maximum height. The time of flight for the projectile is the same for both the upward and downward motion.
Using the equation for vertical motion:
t = (Vy - Vy0) / (-g)
where g is the acceleration due to gravity (approximated as 9.8 m/s²).
Since the shot is released from a height (y0) of 1.89 m, we need to calculate the maximum height (H) reached above this initial height. H = y - y0.
The vertical displacement (y) can be calculated using the equation:
y = y0 + (Vy0 * t) + (0.5 * (-g) * t^2)
Substituting the values and solving these equations will give us the maximum height and the time at which it occurs.
2. Range and Time of Flight:
For the horizontal motion, we can analyze the range (R) and the time of flight (T).
The horizontal velocity (Vx) can be calculated by multiplying the initial speed (v) by the cosine of the angle (θ):
Vx = v * cos(θ)
The range (R) can be found using the equation:
R = Vx * T
where T is the time of flight. Since the horizontal and vertical motions are independent, the time of flight (T) for the projectile can be calculated as twice the time it takes to reach the maximum height.
T = 2 * t
3. Optimum Projection Angle:
The optimum projection angle is the angle at which the range (R) is maximized. To find this angle, you can create a formula for the range (R) in terms of the angle (θ) and then differentiate it with respect to the angle (θ). Setting the derivative equal to zero and solving for θ will give you the optimum projection angle.
To summarize:
1. Find the maximum height reached and the time at which it occurs using the vertical motion equations.
2. Calculate the range and time of flight using the horizontal motion equations.
3. Find the optimum projection angle by differentiating the range equation with respect to the angle (θ) and solving for θ.