91Ó°ÊÓ

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof \(1.00 \mathrm{~s}\) later. Ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{~m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\) and (ii) \(v_{0}\) is \(9.5 \mathrm{~m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\max },\) no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\max }\). The value \(v_{\max }\) has a simple physical interpretation. What is it? (d) If \(u_{0}\) is less than some value \(v_{\min }\), no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\min }\). The value \(v_{\min }\) also has a simple physical interpretation. What is it?

Step by step solution

01

Determine Initial Speed

The first ball is thrown upwards and the second one is dropped 1 second later. For them to land at the same time, the first ball should reach the ground 1 second after it was thrown. Therefore, we can use the equation of motion under gravity to find the initial speed. Given height \(h = 20m\), time \(t = 1s\), and gravity \(g = 9.8m/s^2\), the equation becomes \(20 = v_{0} * 1 - 0.5 * 9.8 * 1^2\). On solving, the initial speed \(v_{0}\) comes out to be \(24.9 m/s\).

02

Determine Building Height

The same equation of motion can be used to find the building height for given initial speeds. For \(v_{0} = 6.0 m/s\), the equation becomes \(h = 6 * 1 + 0.5 * 9.8 * 1^2\), giving \(h = 10.9m\). Similarly for \(v_{0} = 9.5 m/s\), the equation becomes \(h = 9.5 * 1 + 0.5 * 9.8 * 1^2\), giving \(h = 14.65m\).

03

Solve for Maximum Velocity

If the initial speed is such that it makes the square root in the equation of motion imaginary then no real height exists. By equating the discriminant \(v_0^2 - 4 * g/2 * 0\) of the quadratic equation to zero we get \(v_0 = v_{max} = 9.8 m/s\). Anything greater than this will not give a real solution for height. This implies that the ball reaches the maximum height and falls back only after a time greater than 1 second.

04

Solve for Minimum Velocity

If the initial speed is such that it takes more than 1 second to decelerate to zero velocity then no appropriate height exists. The time taken \(t\) to decelerate to zero (when the ball reaches maximum height) is given by \(v_0 = g * t\). By equating this to 1 second we get \(v_0= v_{min} = 9.8 m/s\). Anything less than this means the ball will not have reached its maximum height in 1 second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Projectile Motion

Projectile motion refers to the form of motion experienced by an object that is launched into the air and influenced only by the forces of gravity and its initial launch velocity. In the scenario with the two balls, the first ball being thrown up and the second ball being dropped can both be described by projectile motion principles.

To understand how both balls can hit the ground simultaneously, it is essential to explore the characteristics of projectile motion. It follows a parabolic trajectory and its motion can be broken down into horizontal and vertical components. Since the balls are moving vertically in this problem, we are primarily concerned with the effects of gravity on their vertical motion. The initial speed and direction determine how high and how far they will go before falling back down under the force of gravity.

Equations of Motion

The equations of motion are a set of formulas that allow us to calculate different properties of an object's motion, such as displacement, velocity, and acceleration, at any given time. When considering free fall and vertical projectile motion, the main equation at play is \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the displacement, \( u \) is the initial velocity, \( t \) is the time, and \( a \) is the acceleration due to gravity.

In our exercise, we can rearrange this equation to solve for different variables depending on the given information, such as finding the initial speed of the first ball or the height of the building. The steps provided in the solution utilize this equation adeptly to deduce the required values.

Free Fall

Free fall occurs when an object is falling solely under the influence of gravity, with no propulsive force and ideally no air resistance. Both balls in the given problem experience free fall after they are released; the first ball after reaching its peak height and beginning its descent, and the second ball immediately upon release.

Understanding the concept of free fall is integral to solving kinematics problems involving objects in motion near the Earth’s surface. The acceleration due to gravity, commonly denoted as \( g \), is approximately \( 9.8 m/s^2 \) and acts downwards toward the center of the Earth. The calculation of the building height or the initial speed invariably involves principles of free fall.

Velocity

In physics, velocity is a vector quantity that refers to both the speed of an object and the direction of its motion. For vertical motion problems like the one we are discussing, velocity can be either upward or downward, and its magnitude changes constantly due to the acceleration caused by gravity.

Velocity plays a crucial role in determining motion outcomes. For example, the initial velocity (\( v_0 \)) affects how long the first ball remains in the air and how it eventually synchronizes with the second ball to hit the ground at the same time. The concept of maximum and minimum velocities (\( v_{max} \) and \( v_{min} \)) further provides an insight into the peak limits at which the balls' motions obey the time constraint imposed by the problem's conditions.