Question

A projectile is launched from the top of a building with an initial velocity of \(30 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal. The magnitude of its velocity at \(t=5 \mathrm{~s}\) after the launch is a) \(-23.0 \mathrm{~m} / \mathrm{s}\) c) \(15.0 \mathrm{~m} / \mathrm{s}\) e) \(50.4 \mathrm{~m} / \mathrm{s}\) b) \(7.3 \mathrm{~m} / \mathrm{s}\) d) \(27.5 \mathrm{~m} / \mathrm{s}\)

Short answer

a) 15 m/s b) 20 m/s c) 25 m/s d) 27.5 m/s Answer: d) 27.5 m/s

Step-by-step solution

Identify the given values

We are given the following information: - Initial velocity: \(30\mathrm{~m}/\mathrm{s}\) - Launch angle: \(60^{\circ}\) - Time: \(5\mathrm{~s}\)

Determine the initial horizontal and vertical velocities

To find the initial horizontal and vertical velocities, we can use the following trigonometric formulas: - Horizontal component: \(v_{0x} = v_{0}\cos{(\theta)}\) - Vertical component: \(v_{0y} = v_{0}\sin{(\theta)}\) Using the given values: \(v_{0x} = 30\mathrm{~m}/\mathrm{s} \times \cos{60^{\circ}} = 15\mathrm{~m}/\mathrm{s}\) \(v_{0y} = 30\mathrm{~m}/\mathrm{s} \times \sin{60^{\circ}} = 25.98\mathrm{~m}/\mathrm{s}\)

Calculate the vertical velocity at the given time

We'll now use the kinematic equation: \(v_y = v_{0y} - gt\) where \(v_y\) is the vertical velocity at time \(t\), \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.81\mathrm{~m}/\mathrm{s^2}\)), and \(t\) is the given time (\(5\mathrm{~s}\)). \(v_y = 25.98\mathrm{~m}/\mathrm{s} - (9.81\mathrm{~m}/\mathrm{s^2})(5\mathrm{~s}) = -23.07\mathrm{~m}/\mathrm{s}\)

Calculate the magnitude of the velocity at the given time

Now, we will find the magnitude of the velocity at \(t=5\mathrm{~s}\) using the Pythagorean theorem: \(v = \sqrt{v_{x}^2 + v_{y}^2}\) Since the horizontal velocity remains constant throughout the motion, we have: \(v = \sqrt{(15\mathrm{~m}/\mathrm{s})^2 + (-23.07\mathrm{~m}/\mathrm{s})^2} \approx 27.5\mathrm{~m}/\mathrm{s}\) So the correct answer is: d) \(27.5\mathrm{~m}/\mathrm{s}\)

Key concepts

Initial Velocity

The initial velocity of a projectile is the speed at which it is launched into motion. It plays a crucial role in determining the projectile's path and final position. In physics problems, the initial velocity is often given a magnitude (speed) and a direction (angle). For instance, a projectile launched with an initial velocity of 30 m/s at a 60-degree angle has both horizontal and vertical components.

The horizontal component affects how far the projectile will travel, while the vertical component influences the height and flight time. It's important to understand that these components are independent of each other, except they come together to form the initial velocity vector, which in turn affects the entire trajectory of the projectile.

Launch Angle

The launch angle is the angle at which a projectile is launched relative to the horizontal. In our example, the projectile is launched at a 60-degree angle.

The launch angle is significant because it affects the shape of the projectile's trajectory. A launch angle of 45 degrees, assuming no air resistance, will give the maximum range. Angles smaller than 45 degrees will yield flatter trajectories, while angles greater than 45 degrees will result in trajectories with greater height but shorter range. Adjusting the launch angle changes the balance between the horizontal and vertical components of the initial velocity.

Kinematic Equations

Kinematic equations describe the motion of an object under constant acceleration. They are essential for solving problems involving projectile motion, where acceleration due to gravity is constant.

The kinematic equation used in our example to find the final vertical velocity at time t is:
\( v_y = v_{0y} - gt \)
where
\( v_y \) is the final vertical velocity,
\( v_{0y} \) is the initial vertical velocity,
\( g \) is the acceleration due to gravity (9.81 m/s² on Earth), and
\( t \) is the time elapsed. These equations are invaluable for predicting future motion based on initial conditions.

Trigonometric Functions in Physics

Trigonometric functions such as sine and cosine are used in physics to resolve a vector into its perpendicular components - horizontal and vertical. These functions relate the angles of a triangle to the ratios of its sides.

In the context of projectile motion, we use the cosine to find the initial horizontal velocity (adjacent to the angle) and the sine to determine the initial vertical velocity (opposite to the angle). Thus, initial velocities can be represented as:
\( v_{0x} = v_0 \cdot \cos(\theta) \)
and
\( v_{0y} = v_0 \cdot \sin(\theta) \).
The initial velocity vector can thus be divided into two parts, each impacted differently by the forces acting on the projectile.

Magnitude of Velocity

The magnitude of velocity refers to the speed of a projectile at a given instance, without regard to its direction - essentially, how fast it moves. When we consider projectile motion, we must take into account both horizontal and vertical velocities to find the overall speed.

The total velocity (or speed) at any point is the square root of the sum of the squares of its horizontal and vertical components, as per the Pythagorean theorem:
\( v = \sqrt{v_x^2 + v_y^2} \).
This formula helps us calculate the projectile's speed at any given moment during its trajectory, which is fundamental in problems involving motion.