Question

Use a vertical motion model to find how long it will take for the object to reach the ground. You drop keys from a window 30 feet above ground to your friend below. Your friend does not catch them.

Short answer

The time it will take for the keys to hit the ground is approximately 1.37 seconds.

Step-by-step solution

Identify the given information

The keys are dropped from a height of 30 feet. In this case, we're going to use the acceleration due to gravity as \( g = 32 ft/s^2 \) (standard gravity in customary units).

Apply the equation for time of flight

We apply the time of flight equation which is \( t = \sqrt{(2h/g)} \). Substitute the values \( h = 30\,ft \) and \( g = 32\,ft/s^2 \) into the equation.

Solve the equation

We calculate the square root of the obtained result from the above step to get the total time it takes for the keys to hit the ground.

Key concepts

Time of Flight Equation

Understanding the time of flight equation is essential when analyzing the motion of objects in free fall. When you release an object from a certain height, like dropping keys out of a window, the only force acting on it is gravity. This force causes the object to accelerate downwards.

The time of flight equation, given by the formula \( t = \sqrt{(2h/g)} \), allows us to calculate the time an object takes to hit the ground after being dropped from a height \( h \). Here, \( g \) represents the acceleration due to gravity. To apply this formula, we simply insert the known height and the value of \( g \), which for earth's surface is approximately \( 9.8 m/s^2 \), or \( 32 ft/s^2 \) in customary units if dealing with feet.

For our example, the keys are dropped from 30 feet, so by substituting \( h = 30 \,ft \) and \( g = 32\,ft/s^2 \) into the equation and calculating, we can determine how long it takes for the keys to reach the ground. It is a handy equation not just in theoretical physics but also in practical applications like calculating the time for an object to fall for safety assessments or entertainment purposes like designing drop rides in amusement parks.

Acceleration Due to Gravity

The acceleration due to gravity is a fundamental concept in physics that affects every object on Earth. Denoted often as \( g \) and approximated to \( 9.8 m/s^2 \) (or \( 32 ft/s^2 \) when using customary units), it signifies the rate at which an object will speed up as it falls towards Earth.

This acceleration is constant near the Earth's surface, which simplifies calculations involving vertical motion. In the example of dropping keys from a height, gravity is the sole force in the absence of air resistance, giving the object a consistent acceleration until it impacts the ground.

The value of \( g \) can vary slightly depending on altitude and geographical location due to Earth's non-uniform shape and density. However, in most high school and undergraduate physics problems, we use the standard approximation for simplicity. Understanding this concept is crucial for numerous applications, from predicting the motion of planets to designing infrastructure and vehicles.

Square Root Applications in Physics

The square root appears consistently in physical equations, particularly when dealing with quadratic relationships, such as displacement in uniformly accelerated motion. In our vertical motion example, we see it applied within the time of flight equation.

When we compute \( t = \sqrt{(2h/g)} \), the square root is used to reverse the squaring of time that occurs when deriving the formula from the basic equations of motion. It's essential to remember that while square roots can have both positive and negative values, in physical situations like this, we only consider the positive root because time cannot be negative.

Practical Tip

Always pay attention to units when applying the square root in physics. Consistency is key - ensure that the units of height and acceleration due to gravity match, or the square root will give a nonsensical result. This mathematical operation is a powerful tool; it not only helps in solving motion problems but also provides better understanding in areas such as wave phenomena, energy relationships, and field strength calculations.