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Chapter 2: Problem 99

A ball is thrown vertically downward from the top of a \(36.6-\mathrm{m}\) -tall building. The ball passes the top of a window that is \(12.2 \mathrm{~m}\) above the ground \(2.00 \mathrm{~s}\) after heing thrown. What is the specd of the ball as it passes the top of the window?

Short Answer

Expert verified
The speed of the ball as it passes the top of the window is 22.01 m/s.

Step by step solution

01

Identify Known Values

We have the initial height of the building as 36.6 m, the window's height above the ground as 12.2 m, and the time taken to reach this window height is 2.00 s. The acceleration due to gravity, g, is approximately 9.81 m/s虏 (downward).
02

Calculate Initial Drop Distance

Calculate the distance the ball drops to reach the top of the window: 36.6 m (building height) - 12.2 m (window height) = 24.4 m. This is the distance the ball has traveled in the vertical direction.
03

Use Kinematic Equation for Final Velocity

To find the final speed as the ball passes the window, use the kinematic equation:\[ v = u + gt \]where:- \( u \) is the initial velocity (which we need to find first, if not given explicitly as 0 m/s due to 'thrown' terminology),- \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity),- \( t = 2.00 \text{ s} \) (time).
04

Solve for Final Velocity

Given that the ball was thrown downward, let's assume the ball started with an initial speed \( u \), and solve the equation:\[ 24.4 = ut + \frac{1}{2}gt^2 \]\[ 24.4 = 2u + \frac{1}{2}(9.81)(2^2) \]\[ 24.4 = 2u + 19.62 \]\[ 2u = 24.4 - 19.62 \]\[ 2u = 4.78 \]\[ u = 2.39 \text{ m/s} \]Now, plug \( u = 2.39 \text{ m/s} \) into the first kinematic equation:\[ v = 2.39 + 9.81 imes 2 \]\[ v = 2.39 + 19.62 \]\[ v = 22.01 \text{ m/s} \]

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Key Concepts

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

Projectile motion
Projectile motion occurs when an object is launched into the air and moves under the influence of gravity. In the exercise, the ball is thrown vertically downward, a specific case of projectile motion where the motion is confined to a single vertical direction. Concepts of projectile motion include:
  • Initial Velocity: The speed and direction at which an object is initially thrown. In this context, initial velocity is downward and calculated as part of the problem.

  • Vertical Motion: As gravity is the only force acting, the main focus is on understanding how the ball travels downwards.

  • Kinematic Equations: These equations relate various quantities like velocity, acceleration, time, and displacement to analyze motion. In our exercise, they help determine the speed of the ball as it passes the window.
Free fall
Free fall describes any motion of a body where gravity is the only force acting upon it. Even though the ball in the exercise is thrown initially, once released, it effectively enters a free fall condition.
Understanding this simplifies calculations as the only acceleration acting is due to gravity.
  • Constant Acceleration: In free fall, the acceleration of the object is constant because it鈥檚 only affected by gravity, which is approximated as 9.81 m/s虏.

  • Vertical Displacement: This is calculated to determine how far the ball falls in a given time. It's a key step in calculating its speed at a particular point.

During free fall, the object's speed increases due to gravitational acceleration, a factor important in determining the velocity of the ball as it passes the window.
Acceleration due to gravity
Acceleration due to gravity is essential in analyzing any motion involving free fall or projectile motion. This acceleration impacts how quickly an object speeds up as it falls.
Gravity is always acting downward, and it influences:
  • Speed: As seen in the exercise, the ball's speed increases as it falls. Calculation of speed incorporates gravity's influence over time.

  • Kinematic Equations: These rely heavily on gravity to calculate various motion components, like the final velocity or distance traveled.

  • Uniform Force: Gravity is a constant force, meaning its effect is predictable and can be precisely included in calculations.
In our exercise, gravity's acceleration of 9.81 m/s虏 helps us determine both the initial velocity and speed as the ball passes the window, highlighting its integral role in this and many other kinematics problems.

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Most popular questions from this chapter

Catapulting mush- rooms. Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on onc side of the spore and a film grows on the other side. The spore is bent over by the drop's weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of \(1.6 \mathrm{~m} / \mathrm{s}\) in a \(5.0 \mu \mathrm{m}\) launch; its speed is then reduced to zero in \(1.0 \mathrm{~mm}\) by the air. Using those data and assuming constant accelerations, find the acceleration in terms of \(g\) during (a) the launch and (b) the speed reduction.

A motorcyclist who is moving along an \(x\) axis directed toward the east has an acceleration given by \(a=(6.1-1.2 t) \mathrm{m} / \mathrm{s}^{2}\) for \(0 \leq t \leq 6.0 \mathrm{~s}\). At \(t=0,\) the velocity and position of the cyclist are \(2.7 \mathrm{~m} / \mathrm{s}\) and \(7.3 \mathrm{~m}\). (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between \(t=0\) and \(6.0 \mathrm{~s} ?\)

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what spced docs it hit the ground?

When startled, an armadillo will leap upward. Suppose it riscs \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m} ?\) (c) How much higher does it go?

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a 145 m drop tower. This is an evacuated vertical tower through which, among other possibilities, a 1 -m-diameter sphere containing an experimental package can he dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of \(25 \mathrm{~g}\) as its speed is reduced to zero. Through what distance does it travel during the deceleration?
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