Positive (Upward) Launch: Videos & Practice Problems

In non-symmetrical projectile motion, objects can be launched from different heights, affecting their trajectories. When solving these problems, sketch the paths in the x and y axes, identify target variables, and choose appropriate intervals. For example, when calculating the vertical component of velocity just before impact, use the equation $v^d=v^a+at$. Understanding these principles simplifies calculations and enhances comprehension of motion dynamics.

Solving Non-Symmetrical Upward Launch Problems

In projectile motion, when an object is launched upwards and lands at a different height, the motion is classified as non-symmetrical. This can occur when the object lands at a higher or lower elevation than its launch point. Understanding the trajectory of such launches is crucial, as it involves analyzing both the vertical and horizontal components of motion.

For a launch from a height, the trajectory can be visualized in two parts: the ascent to the maximum height and the descent to the final height. When an object is launched from a height, it first travels upwards, reaches a peak, and then descends to a lower height. Conversely, if it lands at a higher height, the object will ascend and then continue to travel upwards after reaching its peak.

To solve these problems, it is essential to identify key points in the motion, such as the initial height, maximum height, and final height. For example, if a potato is launched from a 20-meter-high cliff, the vertical motion can be analyzed by breaking it down into intervals. The initial velocity at launch can be denoted as \( v_a \), and the vertical component of the velocity just before hitting the ground is what we aim to find, denoted as \( v_{dy} \).

In these scenarios, the vertical acceleration due to gravity is consistently \( a_y = -9.8 \, \text{m/s}^2 \). When the object reaches its maximum height, the vertical component of the velocity at that point is zero (\( v_{by} = 0 \)). This simplifies calculations significantly, as it allows us to use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement:

$$ v_{dy}^2 = v_{by}^2 + 2a_y \Delta y $$

In this equation, \( \Delta y \) represents the vertical displacement from the maximum height to the final height. For instance, if the potato reaches a maximum height of 49.4 meters above the ground, the displacement from the peak to the ground would be negative, indicating a downward movement. Thus, \( \Delta y \) would be -49.4 meters.

Substituting the known values into the equation, we find:

$$ v_{dy}^2 = 0 + 2(-9.8)(-49.4) $$

Calculating this gives:

$$ v_{dy}^2 = 960.16 $$

Taking the square root yields two potential solutions for \( v_{dy} \): \( 31.1 \, \text{m/s} \) (upward) and \( -31.1 \, \text{m/s} \) (downward). Since we are interested in the velocity just before the object hits the ground, the correct answer is the downward velocity:

$$ v_{dy} = -31.1 \, \text{m/s} $$

This analysis illustrates the importance of understanding the components of motion and how to apply kinematic equations to solve for unknown variables in non-symmetrical projectile problems.

You throw a rock off the top of a tall building at an upward angle of 15°. At t=3 s, the rock's horizontal displacement from you is 52m. How high does the rock get above the top of the building?

• A. 1.1 m
• B. 4.6 m
• C. 30 m

Answer: (A)

Height of a Roof

In this problem, we analyze the motion of a ball thrown towards a rooftop, aiming to determine the height of the roof. The ball is launched at an initial velocity of 13 m/s at an angle of 67.4 degrees. The maximum height of the ball's trajectory occurs directly above the edge of the roof, with the ball landing 3 meters away from this edge.

To solve for the height of the roof, we define the vertical displacement from the launch point (point A) to the point where the ball hits the roof (point C) as Δy from A to C. This can be expressed as the sum of two vertical displacements: Δy from A to B (the maximum height) and Δy from B to C (the downward motion to the roof). Thus, we have:

Δy from A to C = Δy from A to B + Δy from B to C.

To find Δy from A to B, we utilize the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement, which is:

v_by² = v_ay² + 2a_yΔy_ab.

At the peak (point B), the final vertical velocity v_by is 0 m/s. The initial vertical velocity v_ay can be calculated as:

v_ay = v_initial * sin(67.4°) = 13 * sin(67.4°) ≈ 12 m/s.

Substituting the known values into the kinematic equation gives:

0 = (12)² + 2(-9.8)(Δy_ab).

Solving for Δy from A to B yields approximately 7.35 meters.

Next, we calculate Δy from B to C. Here, the initial vertical velocity is 0 m/s, and we need to find the time taken for the ball to fall from B to C. We first determine the time using the horizontal motion:

Δx_bc = v_x * t_bc.

Given that the horizontal displacement Δx_bc is 3 meters and the horizontal component of the initial velocity v_x = v_initial * cos(67.4°) ≈ 5 m/s, we find:

3 = 5 * t_bc ⇒ t_bc = 0.6 seconds.

Now, we can find Δy from B to C using the equation:

Δy_bc = v_by * t_bc + 0.5 * a_y * t_bc².

Substituting the known values gives:

Δy_bc = 0 * 0.6 + 0.5 * (-9.8) * (0.6)² ≈ -1.76 meters.

Finally, we combine the two vertical displacements to find the total height of the roof:

Δy from A to C = 7.35 m + (-1.76 m) = 5.59 m.

Thus, the height of the roof is approximately 5.6 meters. This method illustrates how breaking down complex projectile motion problems into manageable intervals can simplify the analysis and lead to accurate results.

Catapult Siege

In this problem, we analyze the trajectory of a stone launched by a catapult towards a castle wall. The stone can either hit the wall while ascending or after reaching its maximum height and descending. Regardless of the specific path, we focus on determining the angle of the stone's velocity at the moment it strikes the wall.

To begin, we establish the components of the stone's velocity: the horizontal component (\(v_{cx}\)) and the vertical component (\(v_{cy}\)). The horizontal velocity remains constant throughout the motion, calculated as:

\(v_{cx} = v_0 \cdot \cos(\theta)\)

Given an initial velocity (\(v_0\)) of 50 m/s and an angle (\(\theta\)) of 56 degrees, we find:

\(v_{cx} = 50 \cdot \cos(56^\circ) \approx 28 \, \text{m/s}\)

Next, we need to determine the vertical component of the velocity (\(v_{cy}\)) at the moment of impact. We use the following kinematic equation, which relates initial velocity, acceleration, and time:

\(v_{cy} = v_{ay} + a_y \cdot t_{ac}\)

Here, \(v_{ay}\) is the initial vertical velocity, calculated as:

\(v_{ay} = v_0 \cdot \sin(\theta) = 50 \cdot \sin(56^\circ) \approx 41.5 \, \text{m/s}\)

The acceleration due to gravity (\(a_y\)) is approximately \(-9.8 \, \text{m/s}^2\), and the time of flight (\(t_{ac}\)) is given as 6 seconds. Plugging in these values, we find:

\(v_{cy} = 41.5 + (-9.8) \cdot 6 \approx -17.3 \, \text{m/s}\)

Now, we can calculate the angle (\(\theta_c\)) of the stone's velocity at the moment of impact using the tangent function:

\(\theta_c = \tan^{-1}\left(\frac{v_{cy}}{v_{cx}}\right) = \tan^{-1}\left(\frac{-17.3}{28}\right)\)

Calculating this gives:

\(\theta_c \approx -31.7^\circ\)

The negative angle indicates that the stone is moving below the horizontal when it strikes the wall. Thus, the correct trajectory is the one where the stone descends upon impact. The final answer for the angle of the stone's velocity at the moment of impact is \(-31.7^\circ\).

Using Single Intervals in Positive Launch Problems

In projectile motion problems, the trajectory can be analyzed by breaking it down into horizontal (x) and vertical (y) components. To solve these problems effectively, it is often beneficial to sketch the trajectory and identify the target variable, which is the quantity you need to find. For instance, when calculating the total time of flight from an initial point to a final point, you can choose different intervals to arrive at the same answer. A common approach is to calculate the time for smaller segments and then sum them, or to use a single interval from the start to the end point, which simplifies the calculations.

Consider a cannon fired from a 40-meter cliff at an initial speed of 100 m/s at an angle of 30 degrees. To find the vertical component of the velocity when it hits the ground, you first determine the initial vertical velocity using the sine function:

$$V_{Ay} = V_A \cdot \sin(\theta) = 100 \cdot \sin(30^\circ) = 50 \, \text{m/s}$$

The vertical acceleration due to gravity is constant at:

$$a_y = -9.8 \, \text{m/s}^2$$

The vertical displacement from the cannon to the ground is negative 40 meters, reflecting the downward direction:

$$\Delta y = -40 \, \text{m}$$

With these values, you can apply the kinematic equation that does not involve time:

$$V_{Dy}^2 = V_{Ay}^2 + 2a_y \Delta y$$

Substituting the known values gives:

$$V_{Dy}^2 = 50^2 + 2(-9.8)(-40)$$

Calculating this results in:

$$V_{Dy} = \sqrt{2500 + 784} = \sqrt{3284} \approx 57.3 \, \text{m/s}$$

Since the projectile is moving downward at the point of impact, the vertical component of the velocity is:

$$V_{Dy} = -57.3 \, \text{m/s}$$

Next, to find the total time of flight from the cannon to the ground, you can use another kinematic equation that relates initial velocity, final velocity, acceleration, and time:

$$V_{Dy} = V_{Ay} + a_y \cdot t$$

Substituting the known values:

$$-57.3 = 50 + (-9.8) \cdot t$$

Rearranging gives:

$$-107.3 = -9.8 \cdot t$$

Solving for time:

$$t = \frac{107.3}{9.8} \approx 10.9 \, \text{s}$$

This example illustrates the effectiveness of using a single interval for calculations in projectile motion, simplifying the process and reducing the potential for errors. By focusing on the initial and final states, you can efficiently determine the necessary variables and solve the problem accurately.

A ball is thrown from the top of a 50-m-tall building with a speed of 40 m/s at an angle of 37° above the horizontal. How far horizontally does the ball travel before hitting the ground?

• A. 101.2 m
• B. 50.6 m
• C. 207 m
• D. 414 m

Answer: (C)

Example 3

In this problem, we analyze the motion of a ball thrown from a building at an angle of 37 degrees above the horizontal. The ball travels a horizontal distance of 24 meters before hitting a window on another building that is lower than the starting height of 50 meters. Our goal is to determine the height of the window above the ground.

To solve this, we recognize that this is a projectile motion problem, where the ball's trajectory can be broken down into horizontal (x) and vertical (y) components. The vertical motion is influenced by gravity, which accelerates the ball downwards at approximately -9.8 m/s².

We start by defining our variables. The initial height of the ball is 50 meters, and we denote the height of the window as h_window. The vertical displacement, Δy, from the launch point to the window can be expressed as:

h_window = h_b + Δy

Here, h_b is the height of the building (50 meters). The ball is in the air for a total time of 3 seconds before it hits the window, which gives us the time interval Δt = 3 seconds.

Next, we need to find the initial vertical velocity component, v_ay, which can be calculated using the initial velocity v_a and the angle of projection:

v_ay = v_a * sin(37°)

However, we first need to determine the initial velocity v_a. We can find this by analyzing the horizontal motion. The horizontal distance traveled is given by:

Δx = v_ax * Δt

Where v_ax is the horizontal component of the initial velocity:

v_ax = v_a * cos(37°)

Substituting the known values, we have:

24 m = v_a * cos(37°) * 3 s

From this, we can solve for v_a:

v_a = 24 m / (3 s * cos(37°)) ≈ 10.01 m/s

Now, we can find the vertical component:

v_ay = 10.01 m/s * sin(37°) ≈ 6 m/s

With the initial vertical velocity known, we can now calculate the vertical displacement Δy using the kinematic equation that does not involve the final velocity:

Δy = v_ay * Δt + 0.5 * a * (Δt)²

Substituting the values:

Δy = 6 m/s * 3 s + 0.5 * (-9.8 m/s²) * (3 s)²

Calculating this gives:

Δy = 18 m - 44.1 m = -26.1 m

This negative value indicates that the ball has fallen 26.1 meters from its original height. To find the height of the window, we substitute back into our earlier equation:

h_window = 50 m + (-26.1 m) = 23.9 m

Rounding this to the nearest whole number, we find that the height of the window is approximately 24 meters above the ground.

Here’s what students ask on this topic:

What is non-symmetrical projectile motion?

Non-symmetrical projectile motion occurs when an object is launched from a height different from where it lands. This results in a trajectory that is not symmetrical. For example, if an object is launched from a cliff and lands on the ground, the motion is non-symmetrical. The key to solving these problems is to sketch the paths in the x and y axes, identify the target variables, and choose appropriate intervals for calculations. Understanding these principles helps simplify the equations and enhances comprehension of the motion dynamics.

How do you solve non-symmetrical upward launch problems?

To solve non-symmetrical upward launch problems, follow these steps: 1) Sketch the trajectory and draw the paths in the x and y axes. 2) Identify the target variable, such as the vertical component of velocity just before impact. 3) Choose the appropriate interval for calculations. For example, if calculating the vertical component of velocity, use the equation $v_d=v_a+a_{y}t$. This approach simplifies the problem and ensures accurate results.

What equations are used in non-symmetrical projectile motion?

In non-symmetrical projectile motion, the key equations used include: 1) $v_{\mathrm{dy}}=v_{\mathrm{ay}}+a_{y}t$ for vertical velocity, 2) $v_{\mathrm{dy}}2=v_{\mathrm{ay}}2+2a_y\mathrm{\Delta y}$ for vertical displacement, and 3) $\mathrm{\Delta y}=v_{\mathrm{ay}}t+0.5a_{y}t2$ for displacement over time. These equations help in calculating various parameters like velocity, displacement, and time.

How do you determine the vertical component of velocity just before impact in a non-symmetrical launch?

To determine the vertical component of velocity just before impact in a non-symmetrical launch, follow these steps: 1) Identify the initial vertical velocity $v_{\mathrm{ay}}$. 2) Use the equation $v_{\mathrm{dy}}2=v_{\mathrm{ay}}2+2a_y\mathrm{\Delta y}$ to find the final vertical velocity $v_{\mathrm{dy}}$. 3) Plug in the values for initial velocity, acceleration due to gravity, and vertical displacement. This will give you the vertical component of velocity just before the object hits the ground.

What is the importance of choosing the right interval in non-symmetrical projectile motion problems?

Choosing the right interval in non-symmetrical projectile motion problems is crucial for simplifying calculations and ensuring accuracy. The interval determines which variables and equations you will use. For example, using the interval from the maximum height to the ground can simplify calculations because the vertical velocity at the peak is zero. This reduces the number of variables and makes the equations easier to solve. Always try to choose an interval that includes known values and simplifies the problem.