Question

A projectile is launched at a \(60^{\circ}\) angle above the horizontal on level ground. The change in its velocity between launch and just before landing is found to be \(\Delta \vec{v}=\vec{v}_{\text {landing }}-\vec{v}_{\text {launch }}=-20 \hat{y} \mathrm{~m} / \mathrm{s}\). What is the initial velocity of the projectile? What is its final velocity just before landing?

Short answer

Answer: The final velocity of the projectile just before landing is given by \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\), where \(v_{0}\) is the initial velocity of the projectile.

Step-by-step solution

Determine the vertical and horizontal components of the change in velocity

The change in velocity is given as \(\Delta \vec{v} = -20\hat{y}\,\text{m/s}\). Since the projectile is launched on level ground and lands on level ground, there is no horizontal change in velocity, so \(\Delta v_x = 0\). The vertical change in velocity is given as \(\Delta v_y = -20\,\text{m/s}\).

Write the projectile motion equations for vertical and horizontal components

We can write the equations for the vertical and horizontal motion components. Horizontal motion: \(v_{x} = v_{0x}\) Vertical motion: \(v_{y} = v_{0y} - gt\) where \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.8\,\text{m/s}^2\)) and \(t\) is the time of flight.

Find the initial vertical velocity component from the change in vertical velocity

Using the equation for vertical motion, we can relate the change in vertical velocity to the initial velocity, \(v_y - v_{0y} = -gt\) Since we know the change in vertical velocity, we can write: \(v_{0y} - (-20\,\text{m/s}) = -gt\) \(v_{0y} = -20\,\text{m/s} + gt\)

Use the launch angle to find the initial horizontal velocity component

We know the projectile is launched at a \(60^{\circ}\) angle above the horizontal, so we can find the initial horizontal velocity component as: \(v_{0x} = v_{0} \cos(60^{\circ})\) Where \(v_{0}\) is the initial velocity of the projectile.

Use the launch angle to find the initial vertical velocity component

Similarly, we can find the initial vertical velocity component using the same angle: \(v_{0y} = v_{0} \sin(60^{\circ})\)

Equating initial vertical velocity components to find the time of flight

From step 3 and step 5, we have, \(v_{0} \sin(60^{\circ}) = -20\,\text{m/s} + gt\) Solve for \(t\): \(t = \frac{v_{0} \sin(60^{\circ}) +20\,\text{m/s}}{g}\)

Find the initial velocity of the projectile

Now, we can square both horizontal and vertical components of initial velocity and sum them to find the initial velocity: \(v_{0}^2 = v_{0x}^2 + v_{0y}^2 \) \(v_{0}^2 = \left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2\) Solve for \(v_{0}\): \(v_{0} = \sqrt{\left(v_{0}\cos(60^{\circ})\right)^2 + \left(v_{0}\sin(60^{\circ})\right)^2}\)

Calculate the final velocity just before landing

To calculate the final velocity just before landing, we account for the fact that the horizontal component remains unchanged throughout the flight and the vertical component has changed as given: \(\vec{v}_{landing} = \left( v_{0}\cos(60^{\circ})\right)\hat{x} + \left(v_{0}\sin(60^{\circ}) -20 \,\text{m/s}\right)\hat{y}\)

Key concepts

Velocity Components

In projectile motion, the velocity of an object can be broken down into two components: horizontal and vertical. These are known as the velocity components and are crucial when analyzing the trajectory of a projectile.

• Horizontal Velocity Component (\(v_{0x}\)): This component remains constant throughout the motion in ideal conditions, meaning no air resistance. It's affected by the initial speed and the angle of launch. Mathematically, it is represented as \(v_{0x} = v_{0} \cos(\theta)\), where \(\theta\) is the launch angle.
• Vertical Velocity Component (\(v_{0y}\)): This component is affected by gravity, causing the projectile to eventually be pulled downward. It's represented as \(v_{0y} = v_{0} \sin(\theta)\).

Understanding these components helps analyze how the angle and speed at launch affect the entire motion of the projectile.

Launch Angle

The launch angle, denoted by \(\theta\), is the angle at which a projectile is launched relative to the horizontal ground. It plays a crucial role in determining the projectile's trajectory, range, and flight time.

• Effect on Range: The range of a projectile is affected by the launch angle. At an angle of \(45^{\circ}\), a projectile achieves the maximum range for a given initial speed without any air resistance.
• Effect on Maximum Height: Higher launch angles result in a steeper trajectory, allowing the projectile to reach a greater maximum height. However, this often reduces the horizontal distance traveled.

By considering the launch angle along with the initial velocity, we can fully determine the motion path and impact point for projectiles.

Time of Flight

Time of flight refers to the total duration the projectile remains in motion from the moment it leaves the launch point to when it returns to the same horizontal level. It is a significant element in projectile motion since it is determined by the vertical displacement and the acceleration due to gravity.

• Vertical Motion Equation: The time of flight can be found using the equation \(v_{0y} = gt\). This equation is derived from the relationship between initial vertical velocity, the acceleration due to gravity, and time.
• Solving for Time: To find the time of flight, rearrange the equation to \(t = \frac{v_{0y} + v_y}{g}\), meaning the time it takes for the initial vertical velocity to scale down to zero and accelerate back to its initial vertical component when impacting the ground.

Utilizing the time of flight helps predict exactly where and when the projectile will land.

Acceleration Due to Gravity

Acceleration due to gravity, represented by the letter \(g\), is a fundamental aspect of projectile motion. Near Earth's surface, this acceleration is approximately \(9.8\,\text{m/s}^2\) directed downwards. It influences how projectiles behave in motion.

• Effect on Vertical Velocity: Gravity steadily decreases the vertical velocity of the projectile during its ascent and equally increases it during descent.
• Air Resistance Neglect: In calculations, gravity is often the only force acting on the vertical component, ignoring air resistance to simplify the situation.

Understanding gravity's role makes it possible to predict how fast and high a projectile will move at any given point in its path.