Question

An outfielder throws a baseball with an initial speed of \(32 \mathrm{~m} / \mathrm{s}\) at an angle of \(23^{\circ}\) to the horizontal. The ball leaves his hand from a height of \(1.83 \mathrm{~m}\). How long is the ball in the air before it hits the ground?

Short answer

Answer: The baseball is in the air for approximately 2.55 seconds before it hits the ground.

Step-by-step solution

1. Calculate the initial vertical velocity

First, we need to determine the initial vertical velocity of the baseball. To do this, we will use the given initial speed and angle to find the vertical component of the velocity using the sine function. The equation for the initial vertical velocity is: \(v_{y0} = v_0 \sin(\theta)\) \(v_{y0} = 32 \mathrm{~m/s} \sin(23^{\circ})\) \(v_{y0} ≈ (32)(0.3907) = 12.5 \mathrm{~m/s}\)

2. Calculate the time to reach the highest point

Next, we need to find the time it takes for the baseball to reach the highest point. To do this, we will use the equation of motion for a projectile in the vertical direction: \(v_y = v_{y0} - g t\) Here, \(v_y\) is the final vertical velocity at the highest point (which is \(0\)), \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)), and \(t\) is the time. We will solve for the time when \(v_y = 0\). \(0 = 12.5 \mathrm{~m/s} - 9.81t\) \(9.81t = 12.5 \mathrm{~m/s}\) \(t = \frac{12.5 \mathrm{~m/s}}{9.81 \mathrm{~m/s^2}}\) \(t ≈ 1.27 \mathrm{~s}\)

3. Calculate the total time in air

Now, we need to find the total time the baseball is in the air. We will use the following equation of motion for a projectile in the vertical direction: \(y = y_0 + v_{y0} t - \frac{1}{2} g t^2\) Here, \(y\) is the final vertical position (which is \(0\)), \(y_0\) is the initial vertical position, \(v_{y0}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. We will solve for the time when \(y = 0\) and \(y_0 = 1.83 \mathrm{~m}\). \(0 = 1.83 \mathrm{~m} + 12.5 \mathrm{~m/s} t - \frac{1}{2}(9.81 \mathrm{~m/s^2})(t^2)\) This is a quadratic equation, so we will use the quadratic formula to solve for \(t\): \(t = \frac{-v_{y0} \pm \sqrt{(v_{y0}^2 - 2gy_0)}}{g}\) Using the calculated value of \(v_{y0}\), we get: \(t = \frac{-12.5 \mathrm{~m/s} \pm \sqrt{(12.5^2 - 2(9.81)(1.83))}}{9.81}\) \(t ≈ -0.149 \mathrm{~s}\) or \(t ≈ 2.55 \mathrm{~s}\) Since the time cannot be negative, we choose the positive value \(t = 2.55 \mathrm{~s}\). The ball is in the air for approximately \(2.55\) seconds before it hits the ground.

Key concepts

Initial Velocity

The concept of initial velocity is fundamental in projectile motion. It refers to the speed at which an object is launched into the air. For our baseball exercise, the initial velocity is given as 32 meters per second. This speed can be broken down into two components: horizontal and vertical velocity.

• **Horizontal Velocity (\(v_{x0}\)):** This is the velocity parallel to the ground.
• **Vertical Velocity (\(v_{y0}\)):** This is the velocity perpendicular to the ground.

We calculate these components using trigonometric functions based on the angle of projection, which helps in determining the object's path and time in the air.

Angle of Projection

The angle of projection is the angle at which the projectile is launched relative to the horizontal. In this example, the angle is 23 degrees. The angle affects how high and how far the object will travel.

• A smaller angle typically results in a flatter flight path.
• A larger angle means a higher arc and potentially more time in the air.

The angle impacts both the vertical and horizontal components of initial velocity. Using trigonometric functions like sine and cosine, we can resolve the initial velocity into these components.

Time of Flight

Time of flight refers to the total time the projectile remains in the air. It's influenced by the initial velocity, the angle of projection, and any initial height from which it is launched. To determine the time of flight in projectile motion, we consider factors like gravity and vertical motion.

• **Upward Motion Time:** Time taken to reach the highest point.
• **Downward Motion Time:** Time taken to fall from the highest point to the ground.

For our baseball scenario, using equations of motion, the total time of flight calculated is approximately 2.55 seconds.

Vertical Motion

Vertical motion describes how the projectile moves up and down under the influence of gravity. Initially, the object moves upwards with an initial vertical velocity, but as gravity acts on it, the velocity decreases, reaches zero at the highest point, and then the object moves downwards.
The formula \(v_y = v_{y0} - g t\) is used to calculate the changing vertical velocity, where \(g\) is the acceleration due to gravity, roughly 9.81 m/s². In the example, the calculations help us determine time to peak height and the total time of flight.

Quadratic Equation in Physics

In physics, quadratic equations are often used to model the parabolic path of projectiles. These equations help calculate key motion variables such as time of flight, maximum height, and range. The general form for a motion equation in a vertical plane is \(y = y_0 + v_{y0} t - \frac{1}{2} g t^2\).
This is a quadratic equation in terms of time (\(t\)).

• In our scenario, solving this using the quadratic formula gave time values, where only the positive time is physically meaningful, leading to finding the duration the baseball stays in the air.
• Quadratic equations demonstrate how initial conditions and gravitational forces combine to dictate projectile trajectories.

Understanding how to solve these equations is crucial for analyzing projectile motion.