Special Equations in Symmetrical Launches: Videos & Practice Problems

In symmetrical projectile motion, the total time of flight can be calculated using the equation tac = 2 × v × sin(θ) / g. The horizontal range is determined by R = v × cosθ⁢t or R = v × 2sin(2θ) / g. Maximum range occurs at a launch angle of 45 degrees, while complementary angles yield the same range. Understanding these principles is crucial for solving projectile motion problems effectively.

Using Special Equations in Symmetrical Launch Problems

In the study of projectile motion, symmetrical launches are characterized by the initial and final heights being equal, leading to specific properties such as equal time ascending and descending, and the upward velocity being the negative of the downward velocity. These properties allow for the use of simplified equations to determine various aspects of the motion.

One key equation for calculating the total time of flight (\(T_{ac}\)) for a symmetrical launch is given by:

\(T_{ac} = \frac{2 v_0 \sin \theta}{g}\)

where \(v_0\) is the initial velocity, \(\theta\) is the launch angle, and \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)). This equation derives from the symmetry of the motion, allowing for a straightforward calculation without needing to separately determine the ascent and descent times.

For example, if a projectile is launched with an initial speed of \(100 \, \text{m/s}\) at an angle of \(53^\circ\), the time of flight can be calculated as:

\(T_{ac} = \frac{2 \times 100 \times \sin(53^\circ)}{9.8} \approx 16.3 \, \text{s}\)

Next, the horizontal range (\(R\)) of the projectile, which represents the total horizontal displacement, can be calculated using the equation:

\(R = v_x \times T_{ac}\)

where \(v_x\) is the horizontal component of the initial velocity, calculated as:

\(v_x = v_0 \cos \theta\)

In this case, \(v_x\) would be:

\(v_x = 100 \cos(53^\circ) \approx 60 \, \text{m/s}\)

Thus, the range can be computed as:

\(R = 60 \times 16.3 \approx 980 \, \text{m}\)

Alternatively, if the time of flight is not calculated, the range can also be determined using the range equation:

\(R = \frac{v_0^2 \sin(2\theta)}{g}\)

For the same initial conditions, this would yield:

\(R = \frac{100^2 \sin(106^\circ)}{9.8} \approx 980 \, \text{m}\)

It is important to note that the range is maximized when the launch angle is \(45^\circ\). Additionally, complementary angles (angles that sum to \(90^\circ\)) will yield the same range for the same initial velocity. For instance, if one angle is \(53^\circ\), the complementary angle would be:

\(\theta_2 = 90^\circ - 53^\circ = 37^\circ\)

Both angles will produce the same range when plugged into the range equation, confirming the relationship between complementary angles and projectile motion.

A frog leaves the ground with a speed of 15 m/s and stays in the air for 2.0 seconds. At what angle did the frog jump?

• A. 40.8°
• B. 9.4°
• C. 19.1°

Answer: (A)

A golf ball is hit at ground level at an angle of 31.9° above the horizontal. Its range is 257 m over a level green. What was the magnitude of the golf ball's initial velocity?

• A. 2807 m/s
• B. 69 m/s
• C. 95 m/s
• D. 53 m/s

Answer: (D)

Here’s what students ask on this topic:

What is the total time of flight equation for symmetrical projectile motion?

The total time of flight for symmetrical projectile motion can be calculated using the equation:

$t$_ac=2v⁢sin⁢(θ)g

where $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. This equation simplifies the process of finding the total time a projectile spends in the air when the initial and final heights are the same.

How do you calculate the horizontal range of a projectile in symmetrical motion?

The horizontal range $R$ of a projectile in symmetrical motion can be calculated using two methods:

1. Using the total time of flight:

$R=v\cos \left(\theta \right)t$_ac

2. Using the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

In both equations, $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. Both methods will yield the same result for the horizontal range.

Why is the range of a projectile maximum at a 45-degree launch angle?

The range of a projectile is maximum at a 45-degree launch angle because this angle optimizes the balance between the horizontal and vertical components of the initial velocity. At 45 degrees, the sine and cosine functions are equal, maximizing the product $\sin \left(2\theta \right)$ in the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

Since $\sin (90°)$ is 1, the range is maximized. Any deviation from 45 degrees results in a smaller product, thus reducing the range.

What are complementary angles in projectile motion, and how do they affect the range?

Complementary angles in projectile motion are two angles that add up to 90 degrees, such as 30 degrees and 60 degrees. These angles yield the same horizontal range for a given initial velocity. This is because the sine of twice the angle is the same for both angles:

$\sin (230°)=\sin (260°)$

Thus, the range equation:

$R=\frac{v^2\sin \left(2\theta \right)}{g}$

yields the same result for both complementary angles, demonstrating that they produce identical ranges.

How do you derive the total time of flight equation for symmetrical projectile motion?

To derive the total time of flight equation for symmetrical projectile motion, start with the vertical motion equation:

$v$_cy=v_ay+a_y⁢t_ac

For symmetrical motion, the final vertical velocity $v$_cy is the negative of the initial vertical velocity $v$_ay:

$v$_cy=-v_ay

Substitute and solve for $t$_ac:

$-v$_ay=v_ay+-g⁢t_ac

$-2v$_ay=-g⁢t_ac

$t$_ac=2v_ayg

Since $v$_ay=v⁢sin⁢(θ), the equation becomes:

$t$_ac=2v⁢sin⁢(θ)g