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Chapter 4: Problem 39

The "hang time" of a punt is measured to be \(4.50 \mathrm{s}\). If the ball was kicked at an angle of \(63.0^{\circ}\) above the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

Short Answer

Expert verified
The initial speed of the ball was approximately 24.77 m/s.

Step by step solution

01

Understanding the Problem

Given the 'hang time' of the punt is 4.50 seconds and the angle of projection is \(63.0^{\circ}\), we need to calculate the ball's initial speed when it's caught at the same level. We will use the formula for the projectile motion.
02

Analyzing Vertical Motion

The vertical motion can be described by the physics formula for time of flight: \[ t = \frac{2v_{0y}}{g} \]where \( t \) is the time of flight (4.50 seconds), \( v_{0y} \) is the initial vertical velocity, and \( g \) is the acceleration due to gravity (\( 9.81 \mathrm{m/s^2} \)).Re-arranging the formula, we get: \[ v_{0y} = \frac{gt}{2} \]
03

Calculating Initial Vertical Velocity

Substitute \( g = 9.81 \mathrm{m/s^2} \) and \( t = 4.50 \mathrm{s} \) into the equation:\[ v_{0y} = \frac{9.81 \times 4.50}{2} = 22.0725 \mathrm{m/s} \].
04

Finding Total Initial Speed

Since \[ v_{0y} = v_0 \sin(\theta) \],where \( \theta = 63.0^{\circ} \), the initial vertical velocity \( v_{0y} \) is \[ v_{0y} = v_0 \sin(63^{\circ}) \].Substitute \( v_{0y} = 22.0725 \mathrm{m/s} \) and solve for \( v_0 \):\[ v_0 = \frac{22.0725}{\sin(63^{\circ})} \].
05

Calculate Initial Speed

Using a calculator, find \( \sin(63^{\circ}) \approx 0.8910 \). Thus,\[ v_0 = \frac{22.0725}{0.8910} = 24.77 \mathrm{m/s} \].

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Key Concepts

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

Initial Speed Calculation
When dealing with projectile motion, calculating the object's initial speed is crucial. Imagine a ball being punted into the air, following an arc before it lands. The initial speed determines how fast the ball is launched both vertically and horizontally. This speed, denoted as \( v_0 \), combines both components into one overall velocity. To find this speed, especially when the "hang time" or total time spent in the air is known, involves breaking the velocity into vertical and horizontal components.
  • Start by identifying the components: vertical \( (v_{0y}) \) and horizontal \( (v_{0x}) \).
  • The given exercise provides a hang time of 4.50 seconds and a projection angle of \( 63.0^{\circ} \).
  • By finding the vertical component using physics formulas, we can approximate the overall initial speed.
Understanding these components not only helps in solving problems but also offers insight into how different forces affect the ball's trajectory.
Vertical Motion Analysis
Vertical motion analysis is a fundamental part of projectile motion. It involves understanding how the vertical component of motion, acting under gravity, affects overall motion. In this exercise, we analyze how the ball's vertical velocity changes over time, especially using the "hang time" concept. The time it takes for the ball to go up and come back down informs the vertical component.Let's look at some important points:
  • We use the formula for time of flight: \( t = \frac{2v_{0y}}{g} \) where \( g = 9.81 \mathrm{m/s^2} \).
  • This formula helps find the initial vertical velocity \( v_{0y} \), by rearranging to \( v_{0y} = \frac{gt}{2} \).
  • With a total time of 4.50 seconds, substituting values gives \( v_{0y} = 22.0725 \text{ m/s} \).
These calculations hinge on understanding that the upward and downward journeys of the ball are symmetrical, which is why the motion analysis covers the entire flight duration.
Time of Flight
The time of flight refers to the total time an object stays in the air during projectile motion. This factor is pivotal in determining other motion parameters. In our scenario, the punted ball with a "hang time" of 4.50 seconds, travels up and back down to its starting level.Here's a quick guide to understanding its relevance:
  • The time of flight is affected mainly by the vertical velocity component and gravity.
  • A longer time of flight implies a greater initial vertical speed or lower gravity force acting on the object.
  • Using the formula \( t = \frac{2v_{0y}}{g} \), you can compute the vertical component and thus the initial speed.
By using the time of flight, not only can the initial speed be found, but one can also predict the maximum height and range of the object's trajectory.

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Most popular questions from this chapter

A ball rolls off a table and falls \(0.75 \mathrm{m}\) to the floor, landing with a speed of \(4.0 \mathrm{m} / \mathrm{s}\). (a) What is the acceleration of the ball just before it strikes the ground? (b) What was the initial speed of the ball? (c) What initial speed must the ball have if it is to land with a speed of \(5.0 \mathrm{m} / \mathrm{s} ?\)

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of \(4.3 \mathrm{m} / \mathrm{s}\) at an angle of \(15^{\circ}\) below the horizontal. It is released \(0.80 \mathrm{m}\) above the floor. What horizontal distance does the ball cover before bouncing?

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of \(22 \mathrm{m} / \mathrm{s},\) directly toward point \(\mathrm{A}\) (Figure \(4-13) .\) When the ball reaches the second baseman \(0.45 \mathrm{s}\) later, it is caught at point \(\mathrm{B}\). (a) How far were you from the second baseman? (b) What is the distance of vertical drop, AB?

When the twin Mars exploration rovers, Spirit and Opportunity, set down on the surface of the red planet in January of 2004 their method of landing was both unique and elaborate. After initial braking with retro rockets, the rovers began their long descent through the thin Martian atmosphere on a parachute until they reached an altitude of about \(16.7 \mathrm{m}\). At that point a system of four air bags with six lobes each were inflated, additional retro rocket blasts brought the craft to a virtual standstill, and the rovers detached from their parachutes. After a period of free fall to the surface, with an acceleration of \(3.72 \mathrm{m} / \mathrm{s}^{2}\), the rovers bounced about a dozen times before coming to rest. They then deflated their air bags, righted themselves, and began to explore the surface. Figure \(4-25\) shows a rover with its surrounding cushion of air bags making its first contact with the Martian surface. After a typical first bounce the upward velocity of a rover would be \(9.92 \mathrm{m} / \mathrm{s}\) at an angle of \(75.0^{\circ}\) above the horizontal. Assume this is the case for the problems that follow. How far does a rover travel in the horizontal direction between its first and second bounces? A. \(13.2 \mathrm{m}\) B. \(49.4 \mathrm{m}\) C. \(51.1 \mathrm{m}\) D. \(98.7 \mathrm{m}\)

Two youngsters dive off an overhang into a lake. Diver 1 drops straight down, and diver 2 runs off the cliff with an initial horizontal speed \(v_{0}\). (a) Is the splashdown speed of diver 2 greater than, less than, or equal to the splashdown speed of diver \(1 ?\) (b) Choose the best explanation from among the following: I. Both divers are in free fall, and hence they will have the same splashdown speed. II. The divers have the same vertical speed at splashdown, but diver 2 has the greater horizontal speed. III. The diver who drops straight down gains more speed than the one who moves horizontally.
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