Question

An object is projected with initial velocity of \(100 \mathrm{~ms}^{-1}\) and angle of 60 . Find the vertical velocity when its horizontal displacement is \(500 \mathrm{~m} .\left(\mathrm{g}=10 \mathrm{~ms}^{-1}\right)\) (A) \(13.35 \mathrm{~ms}^{-1}\) (B) \(-13.35 \mathrm{~ms}^{-1}\) (C) \(-8.65 \mathrm{~ms}^{-1}\) (D) \(98 \mathrm{~ms}^{-1}\)

Short answer

The final vertical velocity of the object when its horizontal displacement is 500 meters is approximately \(-13.4 \mathrm{~ms}^{-1}\), which is closest to option (B) \(-13.35 \mathrm{~ms}^{-1}\).

Step-by-step solution

Find the initial horizontal and vertical components of the velocity

We are given an initial velocity of \(100 \mathrm{~ms}^{-1}\) and an angle of 60 degrees. To determine the horizontal and vertical components of the velocity, we can use the initial velocity and angle with the following trigonometric relationships: \[V_x = V_i\cos\theta\] \[V_y = V_i\sin\theta\] Using the given initial velocity and angle, we can find the initial horizontal and vertical components: \[V_x = 100\cos(60^\circ) = 100\left(\frac{1}{2}\right) = 50 \mathrm{~ms}^{-1}\] \[V_y = 100\sin(60^\circ) = 100\left(\frac{\sqrt{3}}{2}\right) = 50\sqrt{3} \mathrm{~ms}^{-1}\] Initial horizontal velocity (\(V_x\)) is \(50 \mathrm{~ms}^{-1}\) and initial vertical velocity (\(V_y\)) is \(50\sqrt{3} \mathrm{~ms}^{-1}\).

Find the time taken to reach 500 meters horizontal displacement

To find the time taken for the object to reach a horizontal displacement of 500 meters, we can use the formula for horizontal motion: \[x = V_xt\] We are given the horizontal displacement (\(x\)) as 500 meters and the initial horizontal velocity (\(V_x\)) as \(50 \mathrm{~ms}^{-1}\). Now we need to find the time (\(t\)). Let's rearrange the horizontal motion formula and solve for \(t\): \[t = \frac{x}{V_x}\] Now we can plug in the values: \[t = \frac{500\mathrm{m}}{50\mathrm{~ms}^{-1}} = 10\mathrm{s}\] The time taken for the object to reach 500 meters of horizontal displacement is 10 seconds.

Find the final vertical velocity

Now that we have determined the time, we can find the final vertical velocity using the formula for vertical motion: \[V_y' = V_y - gt\] Where \(V_y'\) is the final vertical velocity, \(V_y\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is time. We are given \(g\) as \(10 \mathrm{~ms}^{-2}\). Plugging in the values, we get: \[V_y' = 50\sqrt{3} \mathrm{~ms}^{-1} - 10 \mathrm{~ms}^{-2} \cdot 10\mathrm{s}\] \[V_y' = 50\sqrt{3} - 100\] \[V_y' \approx 86.6 - 100\] \[V_y' \approx -13.4 \mathrm{~ms}^{-1}\] The final vertical velocity of the object is approximately \(-13.4 \mathrm{~ms}^{-1}\). So, the correct answer is (B) \(-13.35 \mathrm{~ms}^{-1}\).

Key concepts

Initial Velocity

When we talk about the initial velocity of an object in projectile motion, we're referring to the speed and direction at which the object is launched. This forms the starting point for any analysis of the projectile's path. In the provided exercise, the initial velocity is given as \(100 \mathrm{~ms}^{-1}\) at an angle of 60 degrees from the horizontal. This means the object moves upward and outward from the point of origin in two dimensions.
To analyze projectile motion, it's crucial to break down this initial velocity into components, because it represents a combination of both horizontal and vertical motions. The initial velocity can greatly affect the time of flight, range, and peak height of the projectile. Understanding and calculating it accurately is the foundation for solving projectile motion problems.

Horizontal and Vertical Components

The horizontal and vertical components of initial velocity tell us how much of the overall velocity is directed along the horizontal axis and how much is directed along the vertical axis. This breakdown is essential for calculating further aspects of the projectile's journey.
The horizontal component (\(V_x\)) is obtained using the cosine of the launch angle, given by \(V_x = V_i\cos\theta\). Meanwhile, the vertical component (\(V_y\)) uses the sine, computed by \(V_y = V_i\sin\theta\). In our scenario, these calculations lead to:

• \(V_x = 100\cos(60^\circ) = 50 \mathrm{~ms}^{-1}\)
• \(V_y = 100\sin(60^\circ) = 50\sqrt{3} \mathrm{~ms}^{-1}\)

These components help us understand how the object will move over time, as they remain constant throughout the object's flight when neglecting air resistance.

Trigonometric Relationships

Trigonometric relationships serve as the bridge to decompose the initial velocity into horizontal and vertical components. They enable us to calculate these components using the angle of projection and the initial velocity magnitude. Using sine and cosine functions tied to the angle helps translate the velocity in terms of two perpendicular directions.
The angle not only affects the speed at which the projectile moves horizontally and vertically but also determines the shape of its path. By manipulating the sine and cosine of the launch angle, you can adjust the ratio between horizontal speed and vertical speed, effectively controlling the projectile's trajectory and altitude.

Horizontal and Vertical Motion Equations

Projectile motion can be analyzed using the equations for horizontal and vertical motion. These equations differ slightly because gravity only influences vertical motion.
For horizontal motion, the equation \(x = V_xt\) shows how far the object travels over time, determined purely by the initial horizontal component and time, given no air resistance.
In vertical motion, gravity plays a role and is captured by the equation \(V_y' = V_y - gt\), where \(V_y'\) represents the final vertical velocity. Gravity (\(g\), typically \(9.8\, \mathrm{m/s}^2\) or approximated as \(10\, \mathrm{m/s}^2\) in simpler calculations) acts downwards, reducing the vertical velocity over time. Understanding these motion equations is crucial for predicting the projectile’s behavior at any point during its flight.