Problem 57 - Calculus with Concepts in Calculus

Question

A ball is thrown downward from the roof of a building 30 meters high, with an initial speed of 5 meters per second. a. Find the approximate height of the ball after \(1 / 2\) second and after 1 second. b. Approximately how long does it take for the ball to reach a window 10 meters above the ground?

Step-by-Step Solution

Answer

a. After 1/2 second, height is 31.275 m; after 1 second, it's 30.1 m. b. It takes approximately 2.177 seconds to reach 10 m above the ground.

Step 1: Understand the Motion Equation

We're dealing with motion under gravity. The height \( h \) at time \( t \) is given by the equation: \[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \] where \( h_0 \) is the initial height (30 meters in this case), \( v_0 \) is the initial velocity (5 m/s), and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

Step 2: Calculate Height After 1/2 Second

Substitute \( h_0 = 30 \) m, \( v_0 = 5 \) m/s, \( g = 9.8 \) m/s², and \( t = 0.5 \) s into the equation: \[ h(0.5) = 30 + 5(0.5) - \frac{1}{2} \times 9.8 \times (0.5)^2 \] Simplifying, we find: \[ h(0.5) = 30 + 2.5 - 1.225 = 31.275 \] meters.

Step 3: Calculate Height After 1 Second

Use the same motion equation for \( t = 1 \) s: \[ h(1) = 30 + 5(1) - \frac{1}{2} \times 9.8 \times (1)^2 \] This simplifies to: \[ h(1) = 30 + 5 - 4.9 = 30.1 \] meters.

Step 4: Solve for Time to Reach 10 Meters

Set \( h(t) = 10 \) and solve for \( t \): \[ 10 = 30 + 5t - \frac{1}{2} \times 9.8t^2 \] Rearrange to form a quadratic equation: \[ 4.9t^2 - 5t - 20 = 0 \] Use the quadratic formula where \( a = 4.9 \), \( b = -5 \), and \( c = -20 \): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Simplifying, you get: \[ t = \frac{5 \pm \sqrt{(-5)^2 + 4 \times 4.9 \times 20}}{9.8} \] Solving this, find the positive root: \[ t \approx 2.177 \] seconds.

Key Concepts

Quadratic Equations

Quadratic equations are an essential part of analyzing projectile motion. They are equations of the form:

• \[ ax^2 + bx + c = 0 \]

In the context of our problem, these equations help us find the time at which the ball reaches a specific height. Here, the quadratic equation helps us solve:

• \[ 4.9t^2 - 5t - 20 = 0 \]

This equation results from substituting the known initial height and speed, along with gravitational acceleration, into the general motion equation. This problem showcases how quadratics describe paths influenced by constant acceleration, such as gravity.
The solution involves using the quadratic formula to find the time variable, \( t \). This formula is:

• \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our scenario, finding the precise moment when the ball is at a certain height requires solving this quadratic equation for the real and positive time value.

Kinematics

Kinematics is the branch of classical mechanics that describes the motion of objects. In this exercise, kinematics is applied using the motion equation:

• \[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \]

This equation predicts the height of the ball at any given time \( t \).
The terms in the equation reflect different aspects of the motion:

• \( h_0 \) is the initial height where the journey starts.
• \( v_0t \) accounts for the initial velocity's influence on height over time.
• \( -\frac{1}{2}gt^2 \) captures how gravity progressively reduces the height.

By substituting specific time values into the motion equation, we can calculate exceptionally accurate predictions of the ball's position at those times. For instance, at \( t = 0.5 \) seconds, the result tells us how far the ball has fallen from its starting point.

Gravitational Acceleration

Gravitational acceleration is a key element in projectile motion problems. It is a constant force that accelerates all objects downward at approximately \( 9.8 \text{ m/s}^2 \) near the Earth's surface.
In the exercise, gravitational acceleration plays a crucial role in determining how fast the ball speeds up as it is thrown downward. In our motion equation:

• \(-\frac{1}{2}gt^2 \)

This term shows how gravity affects the ball’s movement, pulling it down faster with each passing moment. It's essential to factor this constant in to predict how quickly or how slowly the ball descends.
Gravitational acceleration is what transforms a simple kinematic analysis into an exploration of projectile motion. By accounting for this, we can determine not only how high or far an object will travel but the time it will take to reach a particular point in its path.