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Chapter 7: Problem 5

A baseball is thrown from the roof of a 22.0 - -tall building with an initial velocity of magnitude 12.0 \(\mathrm{m} / \mathrm{s}\) and directed at an angle of \(53.1^{\circ}\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of \(53.1^{\circ}\) below the horizontal? \((\mathrm{c})\) If the effects of air resistance are are included, will part (a) or (b) give the higher speed?

Short Answer

Expert verified
(a) Speed is approximately 23.98 m/s; (b) same as (a); (c) part (b) yields higher speed with resistance.

Step by step solution

01

Determine Initial Kinetic and Potential Energy

To solve this, we start by determining the initial kinetic energy of the baseball. Since the initial velocity is at an angle, we find the kinetic energy as: \[ KE_i = \frac{1}{2} m v_i^2 \]where \( m \) is the mass and \( v_i = 12.0 \, \text{m/s} \). Potential energy at the height \( h = 22.0 \, \text{m} \) is given by:\[ PE_i = mgh \]where \( g = 9.8 \, \text{m/s}^2 \).
02

Calculate Final Speed Using Conservation of Energy

The total mechanical energy is conserved since air resistance is ignored:\[ KE_i + PE_i = KE_f + PE_f \]where \( KE_f = \frac{1}{2} m v_f^2 \) is the final kinetic energy and \( PE_f = 0 \) as the ball is on the ground. Plugging the values in, we have:\[ \frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2 \]\[ v_f^2 = v_i^2 + 2gh \]\[ v_f = \sqrt{v_i^2 + 2gh} \]
03

Substitute Values For Ball Thrown Above the Horizontal

For part (a), substitute:\[ v_i = 12.0 \, \text{m/s}, \quad g = 9.8 \, \text{m/s}^2, \quad h = 22.0 \, \text{m} \]Calculate:\[ v_f = \sqrt{(12.0)^2 + 2 \times 9.8 \times 22} \]\[ v_f \approx \sqrt{144 + 431.2} \]\[ v_f \approx \sqrt{575.2} \]\[ v_f \approx 23.98 \, \text{m/s} \]
04

Solution for Ball Thrown Below the Horizontal

For part (b), the angle does not affect the speed just before impact since energy conservation considers only initial speed, ground level, and height. Thus the calculated speed remains:\[ v_f \approx 23.98 \, \text{m/s} \] for both cases.
05

Determine the Effect of Air Resistance

Air resistance would cause the ball to lose mechanical energy during flight. The ball with the higher initial vertical component will generally lose more kinetic energy to air resistance. Thus, in real conditions, the ball thrown below will generally have a higher speed just before impact compared to the one thrown above since it spends less time in the air.

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Key Concepts

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

Conservation of Energy
In projectile motion problems like this one, we often use the principle of conservation of energy. This principle states that the total mechanical energy of an object remains constant if there are no non-conservative forces acting on it, like air resistance.

For this baseball problem, initially, the ball has both kinetic energy (due to its velocity) and potential energy (due to its height). We find the kinetic energy using the formula: \[ KE_i = \frac{1}{2} m v_i^2 \] where \( v_i \) is the initial speed.

Potential energy is calculated using:\[ PE_i = mgh \] where \( h \) is the height from which the ball is thrown. At the point just before the ball hits the ground, all its initial potential energy has converted into kinetic energy, since the height \( h \) becomes zero.
Kinetic Energy
Kinetic energy is the energy of motion, and in this scenario, it's one part of the baseball's mechanical energy. When the ball is thrown, it starts with an initial kinetic energy given by:

\[ KE_i = \frac{1}{2} m v_i^2 \] The initial velocity has both horizontal and vertical components, but the kinetic energy formula only worries about the total speed, regardless of direction, as kinetic energy is a scalar quantity.

Just before the ball hits the ground, all its mechanical energy is in kinetic form because potential energy is depleted. The final kinetic energy, \( KE_f \), is responsible for determining the final speed of the baseball. Thus, the energy conversion gives: \[ v_f = \sqrt{v_i^2 + 2gh} \] This expression shows how initial speed and height affect the final speed.
Potential Energy
Potential energy in the context of projectile motion represents energy stored due to the object's position relative to the ground.

The baseball thrown from the building initially has gravitational potential energy: \[ PE_i = mgh \] where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is the height at 22 meters.

As the baseball descends, this potential energy transforms into kinetic energy. At the ground level, when \( h = 0 \), the potential energy becomes zero. The conservation of mechanical energy ensures that it has all converted into kinetic energy, explaining why the potential energy portion diminishes during descent.
Air Resistance Impact
Air resistance is a force that plays a significant role in real-life projectile motion scenarios. However, it complicates calculations since it dissipates some of the mechanical energy as heat, which isn't accounted for in the basic conservation of energy equations.

Ignoring air resistance simplifies the problem by assuming all initial mechanical energy converts directly into kinetic energy at ground level. In reality, though, air resistance slows the ball down, reducing its final speed.

This means when considering air resistance:
  • More time in the air often means more energy loss to air friction.
  • A ball thrown "above the horizontal" stays longer in the air, losing more energy compared to one thrown "below the horizontal."
Therefore, neglecting air resistance might overestimate the speed just before the ball hits the ground.

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Most popular questions from this chapter

CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

Cp A small block with mass 0.0500 kg slides in a vertical circle of radius \(R=0.800 \mathrm{m}\) on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 \(\mathrm{N} .\) What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A )\) . When the spring is released, it projects the block up the incline. At point \(B\) , a distance of 6.00 \(\mathrm{m}\) up the incline from \(A\) . the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

CALE A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible, (a) Calculate the potential-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0 .\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?
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