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Suppose that an object is projected upward with an initial velocity of \(1000 \mathrm{ft} / \mathrm{sec}\) and air resistance is neglected. Find the time to reach maximum height and the maximum height. Ans. \(125 / 4 \mathrm{sec}\); \(15,625 \mathrm{ft}\).

Short Answer

Expert verified
The time to reach maximum height is \(125 / 4\) seconds; the maximum height is \(15,625\) feet.

Step by step solution

01

Understand the motion of the object

The object is projected upwards in a gravitational field with an initial velocity. Without air resistance, the only force acting on the object is gravity, which will decelerate the object at a constant rate equal to the acceleration due to gravity until the object reaches its maximum height, at which point the velocity will be 0.
02

Apply the kinematic equation for velocity

To find the time it takes for the object to reach its maximum height, use the kinematic equation that relates final velocity, initial velocity, acceleration, and time: \(v_f = v_i + at\). Here, \(v_f\) is the final velocity (0 at the maximum height), \(v_i\) is the initial velocity (1000 ft/sec), \(a\) is the acceleration (due to gravity, which is \(-32 \text{ft/sec}^2\) acting downwards), and \(t\) is the time.
03

Calculate the time to reach maximum height

Solve the equation from Step 2 for time \(t\): \(0 = 1000 + (-32)t\). Rearranging the equation and solving for \(t\) gives \(t = -1000 / -32 = 31.25\) seconds, or \(125/4\) seconds.
04

Apply the kinematic equation for displacement

To find the maximum height, use the kinematic equation that relates displacement, initial velocity, acceleration, and time: \(s = v_i t + 0.5 a t^2\). Here, \(s\) is the displacement (maximum height), \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time found in Step 3.
05

Calculate the maximum height

Plug in the values into the displacement equation: \(s = 1000 \times 31.25 + 0.5 \times (-32) \times (31.25)^2\). This gives \(s = 31250 + 0.5 \times (-32) \times 976.5625\), which simplifies to \(s = 31250 - 15625 = 15625\) feet.

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Key Concepts

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

Kinematic Equations
In physics, kinematic equations are formulas that describe the motion of objects without considering the forces that cause this motion. These equations are the foundation for analyzing projectile motion, which consists of two components: vertical and horizontal motion. Since only gravity acts on the projectile in the vertical direction, we use kinematic equations to determine how high and how long it will take for a projectile to reach its maximum height.

For an object projected straight up with an initial velocity but no air resistance, the kinematic equation for velocity (\(v = v_i + at\)) enables us to calculate the time taken to reach the peak of its motion. In this equation,
  • \(v\) is the final velocity (which is 0 at the peak),
  • \(v_i\) is the initial velocity,
  • \(a\) is the acceleration due to gravity (negative because it opposes the motion), and
  • \(t\) is the time.
By knowing the initial conditions, students can find the time to reach maximum height, which is a pivotal step in analyzing projectile motion.
Maximum Height Projectile
The concept of maximum height projectile involves determining the highest vertical position (the apex) a projectile reaches in its flight path. This maximum height is of great interest when solving physics problems, as it represents the point at which the projectile's velocity becomes zero due to the gravitational pull.

Using the kinematic equation for displacement (\(s = v_i t + \frac{1}{2} a t^2\)), where
  • \(s\) represents the displacement, or maximum height,
  • \(v_i\) is the initial velocity,
  • \(a\) is the acceleration due to gravity, and
  • \(t\) is the time calculated previously,
students can solve for the maximum height of a projectile. It's crucial to input the time at which the velocity becomes zero into this equation. This displacement, or the maximum height reached, showcases not only a practical application of kinematic equations but also highlights the symmetrical nature of projectile motion, as the ascent time equals the descent time under the same conditions.
Acceleration due to Gravity
The term acceleration due to gravity is a constant that represents the rate at which objects accelerate towards the Earth when dropped in a vacuum. On Earth's surface, this acceleration is approximately \(9.8 \text{m/s}^2\) or \(32 \text{ft/s}^2\) when using feet and seconds as units.

This constant acceleration is foundational in the study of kinematics and physics as a whole because it's the acceleration experienced by objects under free fall. In projectile motion problems, this value is negative when plugged into kinematic equations because it opposes the upward initial velocity of the projectile. Understanding that gravity is the only force acting on the projectile when air resistance is neglected allows us to predict the object's motion with precision and solve for variables like maximum height and time to reach that height. When students grasp the concept of acceleration due to gravity, they can apply this knowledge to a wide range of physics problems involving projectile motion.

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Most popular questions from this chapter

The population of the earth was about 3 billion in 1970 . It is estimated that the world's population is increasing continuously at a rate of \(2 \%\) per year of the existing population. When will the population of 20 billion be reached?

Show that if a phenomenon obeys the law \(y=e^{k x}\), then for successive values of \(x\) that are in arithmetic progression the corresponding values of \(y\) are in geometric progression. Suggestion: Suppose that the successive values of \(x\) are \(0, h\), \(2 h, \ldots\), and calculate the corresponding values of \(y\).

The more general statement of how an object loses heat, known as Newton's law of cooling, is that the rate of decrease of temperature of an object is continuous and proportional to the difference between the temperature of the object and that of the surrounding medium. Suppose that the surrounding medium is so vast that the heat it absorbs from the object does not appreciably change the medium's temperature so that it remains constant. Suppose that the object is initially at a temperature of \(100^{\circ}\) and loses temperature continuously at the rate of \(1 / 100\) per minute of the difference between its temperature at time \(t\) and the constant temperature \(T_{0}\) of the surrounding medium. Derive the formula that relates the temperature of the object and time. Ans. \(T-T_{0}=\left(100-T_{0}\right) e^{-0.01 t}\).

Suppose that an object loses temperature at the rate of \(0.01\) of the existing temperature not continuously but at the end of each minute. If the temperature is initially \(100^{\circ}\), derive the formula that relates the temperature of the object and the time. Ans. \(T=100(0.99)^{t}\). To fit the physical situation \(t\) can take on only the values \(0,1,2, \ldots\), where the numbers refer to minutes.

Show that \(y=D e^{k x}\) satisfies the equation \(d y / d x=k y\).

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