Problem 43 - Essential Calculus Early Transcendentals

Question

A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, 450 \(\mathrm{m}\) above the ground. (a) Find the distance of the stone above ground level at time \(t .\) (b) How long does it take the stone to reach the ground? (c) With what velocity does it strike the ground? (d) If the stone is thrown downward with a speed of \(5 \mathrm{m} / \mathrm{s},\) how long does it take to reach the ground?

Step-by-Step Solution

Answer

(a) Distance: \( s(t) = 450 - 4.9t^2 \). (b) Time: \(9.6\) s. (c) Velocity: \(94.1\) m/s. (d) \(9.32\) s if thrown at \(5\) m/s.

Step 1: Understanding the Problem

We have a stone being dropped from a height. We're assuming no air resistance and acceleration due to gravity is approximately \(9.8 \, \text{m/s}^2\). Initial velocity for dropping is \(0 \, \text{m/s}\). We need to find the distance at time \(t\), the time to hit the ground, the velocity when it hits the ground, and the time to hit the ground if thrown downward with a speed.

Step 2: Define the Equations of Motion

We use the equation for motion under constant acceleration: \( s = ut + \frac{1}{2} a t^2 \).- Here, \(s\) is displacement, \(u\) is initial speed, \(a\) is acceleration, and \(t\) is time.

Step 3: Solve for Distance as a Function of Time

Set up the equation for distance from the ground: \( s = 450 - \left(0 \, t + \frac{1}{2} \times 9.8 \, t^2 \right) \).Simplify to get:\[ s(t) = 450 - 4.9t^2 \]This represents the height above the ground at any time \( t \).

Step 4: Solve for Time to Hit the Ground

To find the time \(t\) when the stone reaches the ground, set \(s(t)\) to 0:\[ 0 = 450 - 4.9t^2 \]Solve for \(t\): \[ 4.9t^2 = 450 \]\[ t^2 = \frac{450}{4.9} \]\[ t = \sqrt{\frac{450}{4.9}} \approx 9.6 \, \text{seconds} \]

Step 5: Solve for Velocity at Impact

Use the velocity formula \( v = u + at \), with \(u = 0\):\[ v = 0 + 9.8 \times 9.6 \approx 94.1 \, \text{m/s} \]This is the velocity of the stone as it hits the ground.

Step 6: Solve for Time When Thrown Downward

If thrown downward with speed \(5 \, \text{m/s}\), modify the equation \( s = ut + \frac{1}{2} a t^2 \):\[ 450 = 5t + \frac{1}{2} \times 9.8 \times t^2 \]Simplify:\[ 450 = 5t + 4.9t^2 \]Rearrange to form a quadratic equation:\[ 4.9t^2 + 5t - 450 = 0 \]Solve using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\( a = 4.9, \; b = 5, \; c = -450 \)\[ t = \frac{-5 \pm \sqrt{5^2 - 4 \times 4.9 \times (-450)}}{2 \times 4.9} \]Find \( t \approx 9.32 \, \text{seconds} \).

Key Concepts

Equations of Motion

In physics, the equations of motion are mathematical formulations that describe how an object's velocity and position vary with time under the effect of one or more forces. The three primary kinematic equations used universally include:

• \( s = ut + \frac{1}{2} a t^2 \)
• \( v = u + at \)
• \( v^2 = u^2 + 2as \)

These equations are essential in solving problems involving constant acceleration, such as a stone dropped from a height. Here, the acceleration due to gravity is consistent at approximately \(9.8 \, \text{m/s}^2\). These tools allow us to model the stone's path, assess time till impact, and determine impact speed.

The first equation helps find position over time and was crucial in determining how the height changes during the stone's fall. Meanwhile, the second equation helped calculate the stone's speed right before it hits the ground, clarifying both velocity and time parameters. These fundamental formulas allow students to predict and comprehend projectile motion dynamically and are crucial for understanding motion in straight lines.

Free Fall

Free fall describes an object's motion under the sole influence of gravity. When released from a height, as with the stone in the original exercise, the object experiences gravitational acceleration. In this scenario, air resistance is negligible, and the stone's acceleration remains constant at \(9.8 \, \text{m/s}^2\).

The stone’s initial velocity was \(0 \, \text{m/s}\) since it was simply dropped instead of being pushed downward. As time progresses, the stone accelerates, increasing its velocity until it contacts the ground. Understanding free fall is fascinating as it shows how gravity works straightforwardly, without interference from other forces. Analyzing its path helps us securely grasp other concepts of motion affected differently by added forces or starting velocities.

Quadratic Equation

Quadratic equations, featuring the general form \( ax^2 + bx + c = 0 \), are pivotal in situations like our stone being thrown. Solving quadratic equations allows us to find unknown variables like time when relying on the modified equations of motion. In our scenario, altering the initial conditions by throwing the stone downward changed the dynamics of the equation.

We applied the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to determine the exact time when the stone reached the ground, factoring in initial velocity of \(5 \, \text{m/s}\). It provided a precise solution critical for analyzing real-life applications where initial speed alters the trajectory and outcome noticeably. This illustrates the great utility of quadratic equations beyond pure mathematics, connecting them tightly to physical phenomena like motion studies.

Kinematics

Kinematics, a branch of mechanics, is dedicated to analyzing motion without delving into the forces that cause it. Our exercise embodies kinematic principles by examining motion parameters such as displacement, velocity, and time based on the stone's free-falling motion.

• Displacement: How far the stone travels vertically from the height from which it is dropped.
• Velocity: The changing speed of the stone as it falls down due to gravity.
• Time: The duration it takes to reach the ground or other specified points in its path.

These domains together help create a thorough understanding of how objects behave when in motion. In this problem, examining each of these kinematic attributes demystifies the motion characteristics of the falling stone, emphasizing the link between time, velocity, and displacement pivotal in real-world application.