Question

Use \(s=v_{0} t+\frac{1}{2} g t^{2}\) for these falling-body problems, but be careful of the signs. If you take the upward direction as positive, \(g\) will be negative. An object is thrown upward with an initial speed of \(120 \mathrm{m} / \mathrm{s}\). Find the time for it to return to its starting point. (In the metric system, \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) )

Short answer

The object returns to its starting point after approximately 24.49 seconds.

Step-by-step solution

Convert units for consistency

First, ensure all units are consistent. Convert the acceleration due to gravity, given in centimeters per second squared, to meters per second squared. Since there are 100 centimeters in a meter, divide 980 cm/s^2 by 100 to get 9.8 m/s^2 (make sure to take it as negative because upward direction is positive).

Understand the problem setup

You need to find the time it takes for the object to return to the starting point. At the starting point, the displacement () is zero. So, the equation of motion becomes 0 = v_0 t - (1/2) g t^2, with upward direction as positive.

Apply the equation of motion

Substitute the known values into the equation: 0 = 120t - (1/2) * 9.8 * t^2.

Simplify and solve the quadratic equation

Rearrange the equation to a standard quadratic form: (1/2) * 9.8 * t^2 - 120t = 0. Factor out t: t((1/2) * 9.8 * t - 120) = 0.

Find the non-zero solution for time

The equation t((1/2) * 9.8 * t - 120) = 0 has two solutions: t = 0 (at the initial throw) and t = (120*2)/9.8. Calculate the non-zero solution to find the time when the object returns to its starting point.

Key concepts

Kinematic Equations

Kinematic equations are a set of formulas used to describe the motion of objects with constant acceleration, without the influence of forces. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In our discussed problem, the kinematic equation we use is \( s = v_{0}t + \frac{1}{2}gt^2 \).

When an object is propelled upwards, gravity will decelerate it until it comes to a stop at the peak of its motion. It will then accelerate back downwards until it reaches its starting point. This motion is symmetrical, meaning the time taken to go up is the same as the time to come down, and this principle is crucial when using the given equation.

In the step-by-step solution provided, we look for the time it takes the object to return to its starting point, which means the displacement (\(s\)) is zero. The kinematic equation represents a parabola on a displacement-time graph, and the task is essentially finding the time at which this parabola intersects the time axis. This is where the kinematic equation simplifies to a quadratic equation, which offers two points in time: the moment the object is released and when it returns to the starting height.

Quadratic Equations

Quadratic equations come in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations can have zero, one, or two real solutions, which we can find by factoring, completing the square, or using the quadratic formula. In the context of falling body problems, the quadratic equation represents the points in time where the object is at a certain height.

For our specific problem, after initial rearrangement, we arrive at the quadratic equation \( \frac{1}{2} \cdot 9.8 \cdot t^2 - 120t = 0 \). Factoring an \( t \) out gives us:\( t(\frac{1}{2} \cdot 9.8 \cdot t - 120) = 0 \).

This equation is set to zero because we are interested in the time at which the object's displacement (\(s\)) returns to zero, meaning it's back at the starting height. There are two solutions: \( t = 0 \), which is the time of launch, and \( t = \frac{240}{9.8} \), which is the time of return. We disregard the zero since we are interested in the non-zero value, leading to the final result after the calculation.

Unit Conversion

Unit conversion is the process of converting the value of a physical quantity from one unit to another. This is a crucial step in many physics problems, as equations typically require consistency in units. In our falling body problem, we convert the acceleration due to gravity from centimeters per second squared (\( \mathrm{cm/s^2} \)) to meters per second squared (\( \mathrm{m/s^2} \)). Because 100 centimeters make one meter, we divide 980 \( \mathrm{cm/s^2} \) by 100 to get 9.8 \( \mathrm{m/s^2} \).

It is essential to be mindful of the signs during unit conversion, especially when dealing with vectors like velocity and acceleration. Since an upward direction is considered positive, the acceleration due to gravity must be negative when substituting into the kinematic equation. Proper unit conversion helps avoid errors that could lead to incorrect calculations and results, which is why it is always addressed before plugging values into formulas.