Question

An outfielder throws a baseball with an initial speed of \(32.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(23.0^{\circ}\) 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 remains in the air for approximately 1.37 seconds before hitting the ground.

Step-by-step solution

Understand given variables

The problem provides us with: Initial speed of the baseball: \(v_0 = 32.0 \mathrm{~m/s}\) Angle of throw: \(\theta = 23.0^{\circ}\) Initial height: \(h = 1.83 \mathrm{~m}\) We need to find the time \(t\) the ball stays in the air.

Break down the initial velocity into horizontal and vertical components

We can break down the initial velocity into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components using trigonometry. \(v_{0x} = v_0 \cos(\theta) = 32.0 \cos(23.0^{\circ})\) \(v_{0y} = v_0 \sin(\theta) = 32.0 \sin(23.0^{\circ})\)

Calculate the time taken for the baseball to hit the ground

To find the time taken for the baseball to hit the ground, we look at the vertical motion. Using the kinematic equation: \(y = v_{0y}t - \frac{1}{2}gt^2\), where \(y\) is the vertical displacement, \(t\) is the time, and \(g\) is the acceleration due to gravity (\(g = 9.81 \mathrm{~m/s^2}\)). The ball hits the ground when \(y = h - 1.83 \mathrm{~m}\). Substituting the variables, we get: \(1.83 = v_{0y}t - \frac{1}{2}(9.81)t^2\)

Solve for time

Use the quadratic formula to solve for \(t\): \(t = \frac{-v_{0y} \pm \sqrt{v_{0y}^2 + 2gh}}{g}\). Use only the positive root for the time as it represents the time taken for the baseball to hit the ground (negative time doesn't make sense in this scenario). Calculate the value of \(t\) using the obtained values for \(v_{0y}\) and \(h\). After calculation, the time taken for the baseball to hit the ground is approximately \(t \approx 1.37 \mathrm{~s}\).

Key concepts

Kinematic Equations

Kinematic equations are fundamental tools in physics that describe the motion of objects with constant acceleration. They allow you to solve for various quantities, such as displacement, velocity, time, and acceleration, given enough initial information. In the case of projectile motion, gravity is the constant acceleration acting on the object, typically on Earth with a value of \( 9.81 \mathrm{~m/s^2} \).

To calculate the time a projectile is in the air, we must consider both the horizontal and vertical components of its movement. However, the kinematic equations mainly focus on the vertical displacement since the horizontal motion does not affect the time in the air when neglecting air resistance.

The textbook problem you're working on requires an understanding of the following equation, which is one of the core kinematic equations for vertical motion:
\[ y = v_{0y}t - \frac{1}{2}gt^2 \]
Here, \( y \) represents the vertical displacement, \( v_{0y} \) is the initial vertical velocity, \( t \) is the time the projectile is in the air, and \( g \) is the acceleration due to gravity. Solving this equation for \( t \) involves identifying the initial vertical velocity, the height from which the projectile is thrown, and using the acceleration due to gravity.

Angle of Throw

The angle of throw, or launch angle, is a critical factor determining a projectile's trajectory and the time it remains in the air. When a projectile, like a baseball, is thrown, the initial velocity must be broken down into two components: horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)).

Mathematically, this can be done through basic trigonometry as follows:
\[ v_{0x} = v_0 \cos(\theta) \]
\[ v_{0y} = v_0 \sin(\theta) \]
Here, \( v_0 \) is the initial speed, and \( \theta \) is the angle of throw. The higher the angle, the greater vertical component, but the initial horizontal component decreases. Consequently, at a given initial speed, a greater launch angle typically results in a higher peak and more prolonged time in the air.

In the exercise concerning the outfielder's throw, the angle of throw significantly affects the calculations for \( v_{0y} \), which in turn influence the time the ball is airborne. Understanding this concept is essential for correctly solving problems related to projectile motion, especially when they involve finding the time of flight.

Vertical Displacement

Vertical displacement in projectile motion refers to the change in height of the projectile as it moves through its trajectory. For an object projected upwards, like our baseball, vertical displacement is affected by the initial vertical velocity and the force of gravity pulling the object back down to the ground.

When the baseball is in the air, it will reach a maximum height before descending back to its landing point. The vertical displacement at any point in time during the flight can be calculated using the kinematic equation already discussed. However, for the entire trip, the vertical displacement will be the initial height from which the ball was thrown minus the height when it hits the ground (which is zero when landing on the ground).

In the outfielder's problem, we are ultimately looking to find when the vertical displacement is equal to the negative initial height (\( y = -h \)), because that's the point at which the baseball will hit the ground. By setting up the equation with this end condition, we solve for the time of flight. The approach to tackle this includes using the right kinematic equation and ensuring that the signs for displacement take into account the direction of gravitational pull.