Negative (Downward) Launch: Videos & Practice Problems

Projectile motion can involve launching objects downwards, such as throwing a rock from a building. The initial vertical velocity (v_0y) is negative due to the chosen positive direction being upwards. To find the height of the building, use the equation for vertical displacement: Δy = v_0yt + 1/2a_yt², where a_y is the acceleration due to gravity (g = -9.8 m/s²). The absolute value of the displacement gives the building's height.

Solving Downward Launch Problems

Projectile motion problems can involve launching objects not only upwards or horizontally but also downwards. When an object is thrown downwards, the approach to solving these problems remains consistent with the standard projectile motion equations. The key difference is that the initial vertical velocity component, denoted as \( v_{0y} \), will always be negative due to the convention of considering upward direction as positive.

For instance, consider a scenario where a rock is thrown downwards at an initial speed of 5 meters per second at an angle of 37 degrees from the horizontal. In this case, the angle is measured clockwise from the horizontal, making it effectively negative. The rock lands 10 meters away from the building, and our goal is to determine the height of the building, represented as \( \Delta y \).

To solve this, we first identify the points of interest in the motion: the initial position at the rooftop and the final position when the rock hits the ground. The horizontal displacement \( \Delta x \) is given as 10 meters, while the vertical displacement \( \Delta y \) is what we need to find. The vertical acceleration \( a_y \) is equal to the acceleration due to gravity, which is approximately \( -9.8 \, \text{m/s}^2 \).

Next, we need to break down the initial velocity into its components. The horizontal component \( v_{0x} \) can be calculated using the cosine function, while the vertical component \( v_{0y} \) uses the sine function. Thus, we have:

\( v_{0x} = v_0 \cdot \cos(-37^\circ) \) and \( v_{0y} = v_0 \cdot \sin(-37^\circ) \).

Substituting the values, we find:

\( v_{0x} = 5 \cdot \cos(-37^\circ) \approx 4 \, \text{m/s} \)

\( v_{0y} = 5 \cdot \sin(-37^\circ) \approx -3 \, \text{m/s} \)

With \( v_{0y} \) established, we can now determine the time of flight \( \Delta t \) using the horizontal motion equation:

\( \Delta x = v_{0x} \cdot \Delta t \).

Rearranging gives us:

\( \Delta t = \frac{\Delta x}{v_{0x}} = \frac{10 \, \text{m}}{4 \, \text{m/s}} = 2.5 \, \text{s} \).

Now that we have the time of flight, we can use the vertical motion equation to find \( \Delta y \). The appropriate equation that does not require the final velocity is:

\( \Delta y = v_{0y} \cdot \Delta t + \frac{1}{2} a_y \cdot (\Delta t)^2 \).

Substituting the known values:

\( \Delta y = (-3 \, \text{m/s}) \cdot (2.5 \, \text{s}) + \frac{1}{2} \cdot (-9.8 \, \text{m/s}^2) \cdot (2.5 \, \text{s})^2 \).

Calculating this gives:

\( \Delta y = -7.5 \, \text{m} - 30.625 \, \text{m} = -38.125 \, \text{m} \).

Thus, the vertical displacement from the rooftop to the ground is approximately \( -38.1 \, \text{m} \). Since height is always expressed as a positive value, the height of the building is \( 38.1 \, \text{m} \). This example illustrates that while the direction of the initial velocity affects the sign, the fundamental approach to solving projectile motion problems remains unchanged.

Water pours from a spout at the end of a gutter with a speed of 3.5 m/s, where the spout is angled 45° downwards. The magnitude of the water's velocity when it hits the ground is 14 m/s. How high is the spout from the ground?

• A. 18.8 m above the ground
• B. 10.6 m above the ground
• C. 10.0 m above the ground
• D. 9.4 m above the ground

Answer: (D)

Cannon Firing Downward

In this problem, we analyze the motion of a cannonball fired downward at an angle of 49 degrees below the horizontal. To solve for the horizontal distance traveled by the cannonball before it hits the ground, we begin by establishing a coordinate system with the x-axis representing horizontal motion and the y-axis representing vertical motion.

The initial velocity of the cannonball is given as 73 m/s. To find the components of this velocity, we use trigonometric functions. The horizontal component, \( v_{x} \), is calculated using the cosine function:

\( v_{x} = v_{0} \cdot \cos(-49^\circ) \)

Substituting the values, we find:

\( v_{x} \approx 73 \cdot 0.6561 \approx 47.9 \, \text{m/s} \)

For the vertical component, \( v_{y} \), we use the sine function:

\( v_{y} = v_{0} \cdot \sin(-49^\circ) \)

Calculating this gives:

\( v_{y} \approx 73 \cdot (-0.7547) \approx -55.1 \, \text{m/s} \)

The negative sign indicates that the initial velocity is directed downward.

Next, we need to determine the time of flight, \( t \), to find the horizontal distance. The vertical displacement, \( \Delta y \), is given as -300 m (since the cannonball is falling 300 m). We can use the kinematic equation that relates displacement, initial velocity, acceleration, and time:

\( \Delta y = v_{y} \cdot t + \frac{1}{2} a_{y} \cdot t^2 \)

Here, \( a_{y} \) is the acceleration due to gravity, approximately -9.8 m/s². Plugging in the known values, we have:

\( -300 = -55.1 \cdot t + \frac{1}{2} (-9.8) \cdot t^2 \)

Rearranging this equation leads to:

\( -4.9 t^2 - 55.1 t + 300 = 0 \)

This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \), where \( a = -4.9 \), \( b = -55.1 \), and \( c = 300 \). We can solve for \( t \) using the quadratic formula:

\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substituting the values gives:

\( t = \frac{55.1 \pm \sqrt{(-55.1)^2 - 4 \cdot (-4.9) \cdot 300}}{2 \cdot (-4.9)} \)

Calculating the discriminant and solving yields two potential solutions for \( t \): approximately -15 seconds and 4.0 seconds. Since time cannot be negative, we take \( t = 4.0 \) seconds.

Now, we can find the horizontal distance \( \Delta x \) using the equation:

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

Substituting the values gives:

\( \Delta x = 47.9 \cdot 4.0 \approx 191.6 \, \text{m} \)

Rounding this result, we find that the horizontal distance traveled by the cannonball before it hits the ground is approximately 192 meters.

Here’s what students ask on this topic:

What is the initial vertical velocity in a downward launch?

In a downward launch, the initial vertical velocity (v_0y) is negative because the chosen positive direction is upwards. For example, if an object is thrown downward at an angle, you can calculate v_0y using the equation:

$v_{0}y=v_{0}sin(\theta )$

where θ is the angle of launch measured from the horizontal. If θ is negative (indicating a downward direction), v_0y will also be negative.

How do you calculate the height of a building in a downward launch problem?

To calculate the height of a building in a downward launch problem, you use the vertical displacement equation:

$\mathrm{\Delta y}=v_{0}yt+\frac{1}{2}{a}_y{t}^2$

Here, $v_{0}y$ is the initial vertical velocity, $a_y$ is the acceleration due to gravity (−9.8 m/s²), and $t$ is the time of flight. The absolute value of $\mathrm{\Delta y}$ gives the height of the building.

Why is the initial vertical velocity always negative in a downward launch?

The initial vertical velocity (v_0y) is always negative in a downward launch because the positive direction is defined as upwards. When an object is launched downward, its initial velocity vector points in the opposite direction of the positive y-axis. Therefore, v_0y is negative to reflect this downward direction.

What equations are used to solve downward launch problems?

To solve downward launch problems, you use the same kinematic equations as for any projectile motion. Key equations include:

1. $\mathrm{\Delta y}=v_{0}yt+\frac{1}{2}{a}_y{t}^2$

2. $\mathrm{\Delta x}=v_{0}xt$

3. $v_y=v_{0}y+a_{y}t$

Here, $v_{0}y$ and $v_{0}x$ are the initial velocities in the y and x directions, respectively, and $a_y$ is the acceleration due to gravity.

How do you determine the time of flight in a downward launch problem?

To determine the time of flight in a downward launch problem, you can use the horizontal motion equation:

$\mathrm{\Delta x}=v_{0}xt$

Here, $\mathrm{\Delta x}$ is the horizontal displacement, and $v_{0}x$ is the initial horizontal velocity. Rearrange the equation to solve for time:

$t=\frac{\mathrm{\Delta x}}{v_0}$

Once you have the time, you can use it in the vertical motion equations to find other variables.