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Chapter 2: Problem 45

With what speed must a ball be thrown vertically from ground level to rise to a maximum height of \(50 \mathrm{~m}\) ? (b) How long will it be in the air? (c) Sketch graphs of \(y, v\), and \(a\) versus \(t\) for the ball. On the first two graphs, indicate the time at which \(50 \mathrm{~m}\) is reached.

Short Answer

Expert verified
The ball must be thrown at 31.3 m/s; it is in the air for 6.4 seconds.

Step by step solution

01

Understand the Problem

We are given a problem where we need to determine the speed at which a ball must be thrown upwards to reach a certain height. Additionally, we need to find how long the ball will be airborne and sketch graphs of its position, velocity, and acceleration with respect to time.
02

Identify the Known Quantities

The maximum height, \( h = 50 \) meters, is given. We also know the acceleration due to gravity, \( g = 9.8 \ \text{m/s}^2 \), which acts downwards (thus negatively if upward motion is considered positive).
03

Using the Kinematic Equation for Velocity

We need the initial velocity \( v_0 \) that allows the ball to reach \( 50 \) meters. Using the equation where final velocity \( v = 0 \) at maximum height: \( v^2 = v_0^2 - 2gh \). Set \( v = 0 \), solve for \( v_0 \): \( 0 = v_0^2 - 2 \times 9.8 \times 50 \).
04

Solve for Initial Velocity

Rearrange the equation to get \( v_0^2 = 2 \times 9.8 \times 50 \), which simplifies to \( v_0^2 = 980 \). Taking the square root gives \( v_0 = \sqrt{980} \approx 31.3 \, \text{m/s} \). The initial upward velocity must be approximately \( 31.3 \, \text{m/s} \).
05

Determine the Total Time in the Air

Using another kinematic equation \( v = v_0 - gt \), and setting \( v = 0 \) at the maximum height, solve for time to reach \( 50 \) meters: \( 0 = 31.3 - 9.8t \). Solve for \( t \), we get \( t = \frac{31.3}{9.8} \approx 3.2 \) seconds to reach the peak. The total time in the air is twice this travel time, as it descends back to the ground, so \( 2 \times 3.2 \approx 6.4 \) seconds.
06

Graphical Representation

Sketch the graphs:1. **Position vs. Time**: Draw a parabola opening downward, peaking at \( t = 3.2 \) sec with \( y = 50 \) m.2. **Velocity vs. Time**: Start at \( v = 31.3 \) m/s and linearly decrease to \( v = 0 \) m/s at the peak (\( t = 3.2 \) sec), turning negative upon descent.3. **Acceleration vs. Time**: Plot a constant line at \( a = -9.8 \, \text{m/s}^2 \).

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Key Concepts

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

Kinematic Equations
Kinematic equations help us understand projectile motion by relating different physical quantities like displacement, velocity, and time. In our scenario, where a ball is thrown upward to reach a maximum height of 50 meters, these equations are essential for determining the initial velocity required and the time the ball is in the air. The key equations used here are:
  • \[ v = v_0 + at \], which calculates the final velocity of an object when the initial velocity, acceleration, and time are known.
  • \[ s = v_0t + \frac{1}{2}at^2 \], which gives the displacement. However, in our problem, we used a rearranged form to find the initial velocity:\[ v^2 = v_0^2 - 2gh \]
These equations can help solve a variety of motion problems once certain physical quantities are known. In projectile motion, they are integral as the lack of horizontal forces allows us to focus on vertical forces and motions.
Acceleration due to Gravity
Gravity is a force acting on all objects near the Earth's surface, causing them to accelerate downward at a constant rate of approximately 9.8 m/s虏. In projectile motion, this acceleration has a significant effect, as it opposes any upward force you may apply, such as throwing a ball. This acceleration is always acting downwards:
  • When the motion is upward, gravity decelerates the object.
  • When the object starts falling back, gravity assists its acceleration downwards.
The constant acceleration of gravity makes calculating projectile motion simpler, as it allows us to use kinematic equations effectively. Understanding this concept helps predict how objects will move under its influence, which is critical in planning the initial conditions for desired projectile paths.
Vertical Motion
Vertical motion occurs when an object moves in a straight line perpendicular to the horizontal plane. In the problem discussed, vertical motion is considered as the ball moves up and comes back down in a straight line. Important aspects of vertical motion in projectile scenarios include:
  • At maximum height, the vertical velocity is zero for an instant before gravity pulls the object downwards.
  • The time for ascent equals time for descent if air resistance is negligible.
  • The path taken is symmetrical, meaning the speed of ascent and descent is equivalent at equivalent heights.
This kind of motion is a crucial component in analyzing projectiles, as understanding it allows us to determine how long an object will stay in the air and the characteristics of its trajectory.
Initial Velocity Calculation
Calculating initial velocity is often the first step in solving projectile motion problems. To find how fast a ball must be thrown vertically to reach a specific height, we use kinematic equations and understand the forces acting on the projectile. For our problem, we use:
  • The known height, which in this case is 50 meters.
  • The gravitational force affecting upward motion, i.e., -9.8 m/s虏.
The primary equation to find initial velocity is:\[ v_0^2 = v^2 + 2gh \]For maximum height, \( v = 0 \) m/s, leading to:\[ v_0^2 = 2gh \]Solving for \( v_0 \) gives the speed required initially. Calculating this correctly ensures a projectile reaches its desired point at its peak, making it fundamental to mastering projectile motion problems.

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Most popular questions from this chapter

A car traveling \(56.0 \mathrm{~km} / \mathrm{h}\) is \(24.0 \mathrm{~m}\) from a barrier when the driver slams on the brakes. The car hits the barrier \(2.00 \mathrm{~s}\) later. (a) What is the magnitude of the car's constant acceleration before impact? (b) How fast is the car traveling at impact?

To test the quality of a tennis ball, you drop it onto the floor from a height of \(4.00 \mathrm{~m}\). It rebounds to a height of \(2.00 \mathrm{~m}\). If the ball is in contact with the floor for \(12.0 \mathrm{~ms}\), (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of \(0.50 \mathrm{~s}\), and the topto- bottom height of the window is \(2.00 \mathrm{~m}\). How high above the window top does the flowerpot go?

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m} ;\) both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\) Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2} .\) (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?

When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) 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?
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