Problem 1 - Applied Mathematics

Question

A stone is dropped from the top of a building \(20 \mathrm{~m}\) high. A second stone is dropped from half-way up the same building. Find the time that should elapse between the release of the two stones if they are to reach the ground at the same time.

Step-by-Step Solution

Answer

0.59 seconds

Step 1: Identify the known values and equations

The height of the building is given as 20 m. The first stone is dropped from the top of the building, so its initial height is 20 m. The second stone is dropped from halfway up the building, so its initial height is 10 m. Use the kinematic equation for free fall: \[ h = \frac{1}{2} g t^2 \]where \( h \) is the height, \( g \) is the acceleration due to gravity (approximated as 9.8 m/s^2), and \( t \) is the time.

Step 2: Calculate the time for the first stone

Using the kinematic equation, substitute the height (20 m) and solve for time \( t_1 \):\[ 20 = \frac{1}{2} \times 9.8 \times t_1^2 \]Solving for \( t_1 \) we get:\[ t_1^2 = \frac{40}{9.8} \ t_1 = \frac{\text{sqrt}(40)}{9.8} \approx 2.02 \text{ seconds} \]

Step 3: Calculate the time for the second stone

Using the kinematic equation, substitute the height (10 m) and solve for time \( t_2 \):\[ 10 = \frac{1}{2} \times 9.8 \times t_2^2 \]Solving for \( t_2 \) we get:\[ t_2^2 = \frac{20}{9.8} \ t_2 = \frac{\text{sqrt}(20)}{9.8} \approx 1.43 \text{ seconds} \]

Step 4: Determine the time elapsed between the release of the stones

To ensure the stones reach the ground simultaneously, subtract \( t_2 \) from \( t_1 \):\[ \text{Time elapsed} = t_1 - t_2 = 2.02 - 1.43 = 0.59 \text{ seconds} \]

Key Concepts

Free Fall

When an object is dropped from a height, it undergoes what we refer to as free fall. This movement happens under the influence of gravity alone, without any initial velocity. In our exercise, both stones are in free fall since they are simply dropped and not thrown.
The only force acting on these stones is the gravitational pull of the Earth. Because of this, both stones accelerate downwards at a constant rate, which we'll explain later.
During free fall, there are no air resistance forces considered in our calculations. This means the only factor impacting their descent is gravity.

Acceleration Due to Gravity

The acceleration due to gravity, often represented by the symbol \( g \), is approximately 9.8 m/s^2 near the surface of the Earth. This is a constant value and applies to all objects in free fall, regardless of their mass.
This acceleration rate tells us that for every second an object is in free fall, its velocity increases by 9.8 meters per second. For simplicity, we use this value in our kinematic equations.
In our problem, we use the kinematic equation for free fall: \[ h = \frac{1}{2} g t^2 \] Here, \( h \) represents the height from which the stone is dropped, \( g \) is our constant—9.8 m/s^2, and \( t \) is the time it takes for the stone to reach the ground. By plugging in the respective heights, we can calculate the time for each stone to fall.

Time of Descent

To find out how long it takes for each stone to reach the ground, we use the formula from our kinematic equation. Let's understand this process step by step.
For the stone dropped from 20 meters, we use \[ h = \frac{1}{2} g t^2 \] with \( h = 20 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \). Solving for time, we get \( t_1 \approx 2.02 \text{ seconds} \).
For the stone dropped from 10 meters, using \( h = 10 \text{ m} \), we solve and find \( t_2 \approx 1.43 \text{ seconds} \).
To make sure both stones hit the ground at the same time, we need to delay dropping the second stone by the difference in their times of descent. Calculating this gives us \( 2.02 - 1.43 = 0.59 \text{ seconds} \). Hence, the second stone should be dropped 0.59 seconds after the first stone.