Symmetrical Launch: Videos & Practice Problems

When launching projectiles upwards, the initial velocity is always positive, and the trajectory is symmetrical if the object returns to its original height. The time to reach maximum height can be calculated using the equation $v=v+at$, where the vertical velocity at the peak is zero. For symmetrical launches, the time up equals the time down, and the vertical component of velocity upon return is the negative of the initial vertical velocity, illustrating the conservation of energy and momentum principles.

Solving Symmetric Launch Problems

When analyzing projectiles launched upwards, it's essential to understand that the initial velocity is always positive. For example, when a football is kicked at an angle, its velocity can be broken down into horizontal (x) and vertical (y) components. The vertical component of the initial velocity, denoted as \( v_{0y} \), is positive, while the horizontal component, \( v_{0x} \), is also positive. This upward launch leads to a symmetrical trajectory, meaning that the time taken to reach the maximum height is equal to the time taken to return to the original height.

To solve problems involving symmetrical launches, we can identify key points: the initial point (A), the maximum height (B), and the final point (C) where the projectile returns to the ground. The time taken to reach the maximum height (from A to B) can be calculated using the equation:

\[ v_{by} = v_{ay} + a_y \cdot t_{AB} \]

At the maximum height, the vertical velocity \( v_{by} \) is 0. Given that the acceleration due to gravity \( a_y \) is -9.8 m/s², and the initial vertical velocity \( v_{ay} \) can be calculated using the sine of the launch angle, we can find the time to reach the maximum height. For instance, if \( v_{0} = 20 \) m/s and the angle is 53 degrees, then:

\[ v_{0y} = 20 \cdot \sin(53) \approx 16 \, \text{m/s} \]

Using the equation, we can solve for \( t_{AB} \) and find it to be approximately 1.63 seconds. The total time of flight \( t_{AC} \) is simply double this time due to the symmetry of the trajectory, resulting in a total time of 3.26 seconds.

When the projectile returns to the ground, the vertical component of the velocity \( v_{cy} \) can be calculated using the same equation, but now we consider the entire flight time. Since the vertical displacement from the initial to final position is zero, we can use:

\[ v_{cy} = v_{ay} + a_y \cdot t_{AC} \]

Substituting the known values, we find that \( v_{cy} \) is -16 m/s, indicating that the direction of the velocity is downward. This result illustrates a key principle: for symmetrical launches, the vertical component of the velocity upon return is equal in magnitude but opposite in direction to the initial vertical component. Thus, \( v_{cy} = -v_{ay} \).

In summary, for symmetrical projectile motion, the time to ascend equals the time to descend, and the vertical velocity upon return is the negative of the initial vertical velocity. The overall magnitude of the velocity remains constant throughout the motion, while the angle of the velocity vector changes direction upon descent.

A flare gun launches signal flares with an initial speed of 110 m/s. How far does the flare travel if it is shot at ground level at an angle 65° above the horizontal?

• A. 1890 m
• B. 944 m
• C. 507 m
• D. 1040 m

Answer: (B)

Gravity on Planet X

In this problem, we analyze the motion of a projectile launched on a distant planet, aiming to determine the gravitational acceleration, denoted as \( g \). The projectile is thrown at an initial speed of 10 m/s at an angle of 37 degrees from the horizontal. The trajectory is symmetrical, meaning it will return to the same height from which it was launched.

To solve for \( g \), we first break down the motion into horizontal (x) and vertical (y) components. The initial velocity can be decomposed into its components using trigonometric functions: the horizontal component \( v_{a_x} \) is calculated as \( v_{a} \cos(37^\circ) \) and the vertical component \( v_{a_y} \) as \( v_{a} \sin(37^\circ) \). This gives us \( v_{a_x} = 10 \cos(37^\circ) \approx 8 \, \text{m/s} \) and \( v_{a_y} = 10 \sin(37^\circ) \approx 6 \, \text{m/s} \).

Next, we identify the points of interest in the trajectory: point A (launch), point B (maximum height), and point C (landing). At point B, the vertical velocity \( v_{b_y} \) is 0 m/s. We need to find the vertical displacement \( \Delta y \) from A to B and the time \( t_{ab} \) it takes to reach this height.

Using the horizontal motion, we know the total horizontal distance from A to C is 32 meters. Due to symmetry, the distance from A to B is half of that, or 16 meters. The time to travel this distance can be calculated using the equation:

\( \Delta x = v_{x} \cdot t \)

Substituting the known values, we have:

\( 16 = 8 \cdot t_{ab} \)

Solving for \( t_{ab} \) gives us \( t_{ab} = 2 \, \text{s} \).

Now, we can use the vertical motion equation that relates initial velocity, final velocity, acceleration, and time:

\( v_{b_y} = v_{a_y} + a_y \cdot t_{ab} \)

Substituting the known values, we have:

\( 0 = 6 + a_y \cdot 2 \)

Rearranging gives:

\( a_y = -3 \, \text{m/s}^2 \)

Since \( a_y \) is equal to \(-g\), we find that the magnitude of the gravitational acceleration on this planet is \( g = 3 \, \text{m/s}^2 \). This result indicates that the gravitational force is significantly weaker than that on Earth, where \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

Here’s what students ask on this topic:

What is a symmetrical launch in projectile motion?

A symmetrical launch in projectile motion refers to a scenario where an object is launched upwards and returns to its original height. This results in a parabolic trajectory that is symmetrical about the peak height. The initial vertical velocity is positive, and the time taken to reach the maximum height is equal to the time taken to descend back to the original height. This symmetry simplifies calculations, as the vertical component of velocity upon return is the negative of the initial vertical velocity, and the total time of flight is twice the time to reach the peak.

How do you calculate the time to reach maximum height in a symmetrical launch?

To calculate the time to reach maximum height in a symmetrical launch, you can use the equation:

$v_y=v_{\mathrm{0y}}+a_{y}t$

At the maximum height, the vertical velocity (vy) is zero. Rearranging the equation to solve for time (t):

$t=\; -v_{\mathrm{0y}}/a_y$

Here, v0y is the initial vertical velocity, and ay is the acceleration due to gravity (typically -9.8 m/s²). This will give you the time to reach the maximum height.

What is the total time of flight for a symmetrical launch?

The total time of flight for a symmetrical launch is the time it takes for the projectile to go up to its maximum height and then return to its original height. Since the time to ascend is equal to the time to descend, the total time of flight is:

$T=\; 2t$

where t is the time to reach the maximum height. If t is calculated using $t=\; -v_{\mathrm{0y}}/a_y$, then the total time of flight T is:

$T=\; 2\; (-v_{\mathrm{0y}}/a_y)$

How do you determine the vertical component of velocity upon return in a symmetrical launch?

In a symmetrical launch, the vertical component of velocity upon return to the original height is the negative of the initial vertical velocity. This is due to the symmetry of the motion. If the initial vertical velocity is v0y, then the vertical component of velocity upon return (vcy) is:

$v_{\mathrm{cy}}=\; -v_{\mathrm{0y}}$

This means if the object was launched with an initial vertical velocity of 16 m/s, it will return with a vertical velocity of -16 m/s.

Why is the maximum height a point of interest in symmetrical launches?

The maximum height is a point of interest in symmetrical launches because at this point, the vertical component of the velocity is zero. This simplifies calculations and helps in determining the time to reach the maximum height. Additionally, the maximum height represents the peak of the parabolic trajectory, and understanding this point allows for easier analysis of the projectile's motion. The symmetry around this point also means that the time to ascend to this height is equal to the time to descend from it, making it a crucial aspect of solving symmetrical launch problems.