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Chapter 3: Problem 58

A \(2.00-\mathrm{m}-\mathrm{tall}\) basketball player is standing on the floor \(10.0 \mathrm{~m}\) from the basket, as in Figure P3.58. If he shoots the ball at a \(40.0^{\circ}\) angle with the horizontal, at what initial speed must he throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is \(3.05 \mathrm{~m}\).

Short Answer

Expert verified
To solve for the initial speed, first break down the components of the initial velocity. Use the kinematic equations for the horizontal and vertical motions to form two equations that involve the initial speed and time. Solve these equations simultaneously to get the value of the initial speed.

Step by step solution

01

Components of Initial Velocity

First, break down the initial speed into horizontal and vertical components. The general formulas for these components are: \(V_{0x} = V_{0} \cdot cos(\theta)\) and \(V_{0y} = V_{0} \cdot sin(\theta)\), where \(V_{0}\) is the initial speed and \(\theta\) is the angle of projection.
02

Horizontal Motion Analysis

In the horizontal direction, the motion is uniform since air resistance is normally neglected in these type of problems. So you can use the equation: \(d = V_{0x} \cdot t\), where \(d\) is the horizontal distance (range), \(V_{0x}\) is the horizontal component of initial speed and \(t\) is the time of flight.
03

Vertical Motion Analysis

In the vertical direction, the motion is uniformly accelerated due to gravity. You can use the equation: \(\Delta y = V_{0y} \cdot t - 0.5 \cdot g \cdot t^{2}\), where \(\Delta y\) is the vertical distance (the height difference between the player and the basket), \(V_{0y}\) is the vertical component of initial speed, \(g\) is the acceleration due to gravity and \(t\) is the time of flight.
04

Calculate the Initial Speed

First, substitute \(V_{0x} = V_{0} \cos(\theta)\) into the horizontal equation from step 2, and \(V_{0y} = V_{0} \sin(\theta)\) into the vertical equation from step 3. You will get 2 equations, and both of these have \(t\) and \(V_{0}\) as the variables. Then, you can solve these two equations simultaneously (by equating them) to get the value of \(V_{0}\), the initial speed of the ball.

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Key Concepts

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

Initial Velocity Components
Understanding the initial velocity components is crucial when analyzing projectile motion. In the context of a basketball shot, the player imparts an initial speed to the ball at a certain angle with respect to the horizontal axis. This initial speed, known as the magnitude of the initial velocity, can be broken down into two perpendicular components: horizontal and vertical.

Using trigonometry, we can calculate these components with the equations:
  • \(V_{0x} = V_{0} \times \text{cos}(\theta)\)
  • \(V_{0y} = V_{0} \times \text{sin}(\theta)\)
Here, \(V_{0}\) is the initial speed of the ball, and \(\theta\) represents the angle at which the ball is launched. These components are vital because they allow us to analyze the motion independently along the horizontal and vertical axes, simplifying the overall problem.
Uniformly Accelerated Motion
When we delve into the vertical motion of the projectile, such as a basketball arcing towards the hoop, we are working with uniformly accelerated motion. This means that the only acceleration acting on the ball is due to gravity (\(g\)), and it is constant. In other words, the speed in the vertical direction changes at a steady rate. For the basketball player's shot, the uniform acceleration is directed downwards and has a magnitude of approximately \(9.8 \text{ m/s}^2\).

The key equation for vertically accelerated motion in this scenario is \(\Delta y = V_{0y} \times t - 0.5 \times g \times t^2\). Here, \(\Delta y\) signifies the change in vertical position (height of the basket minus the height of the player's hand), \(V_{0y}\) is the initial vertical component of the velocity, and \(t\) is the time the ball is in the air. Through this equation, we can predict where the ball will be at any given time along its vertical trajectory.
Projectile Motion Equations
The equations of projectile motion allow us to calculate the path of an object, like a basketball, through the air. Since the motion has both horizontal and vertical components, we use separate yet interconnected equations to describe each dimension.

In the horizontal or x-axis, where there is no acceleration once the ball is in the air (assuming no air resistance), we rely on the equation \(d = V_{0x} \times t\), indicative of a uniform motion. Here, \(d\) denotes the horizontal distance the ball needs to travel, \(V_{0x}\) is the initial velocity component along the x-axis, and \(t\) is the time the ball is airborne.

The real challenge occurs when we attempt to solve these equations. They share a common variable, time (\(t\)), which must be consistent for both dimensions. Solving these equations simultaneously gives us the initial speed (\(V_{0}\)) required to ensure the ball travels the correct horizontal distance while also achieving the necessary height to make the shot, without hitting the backboard or undershooting the hoop.

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Most popular questions from this chapter

A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance \(x\) away from the gun. The pellet hits the target a vertical distance \(y\) below the gun. (a) Show that the position of the pellet when traveling through the air is given by \(y=A x^{2}\), where \(A\) is a constant. (b) Express the constant \(A\) in terms of the initial (muzzle) velocity and the freefall acceleration. (c) If \(x=3.00 \mathrm{~m}\) and \(y=0.210 \mathrm{~m}\), what is the initial speed of the pellet?

A hiker starts at his camp and moves the following distances while exploring his surroundings: \(75.0 \mathrm{~m}\) north, \(2.50 \times 10^{2} \mathrm{~m}\) east, \(125 \mathrm{~m}\) at an angle \(30.0^{\circ}\) north of east, and \(1.50 \times 10^{2} \mathrm{~m}\) south. (a) Find his resultant displacement from camp. (Take east as the positive \(x\)-direction and north as the positive \(y\)-direction.) (b) Would changes in the order in which the hiker makes the given displacements alter his final position? Explain.

A man in a maze makes three consecutive displacements. His first displacement is \(8.00 \mathrm{~m}\) westward, and the second is \(13.0 \mathrm{~m}\) northward. At the end of his third displacement he is back to where he started. Use the graphical method to find the magnitude and direction of his third displacement.

BIO A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a speed of \(6.26 \mathrm{~m} / \mathrm{s}\). To move upstream past a waterfall, the salmon does not need to jump to the top of the fall, but only to a point in the fall where the water speed is less than \(3.58 \mathrm{~m} / \mathrm{s} ;\) it can then swim up the fall for the remaining distance. Because the salmon must make forward progress in the water, let's assume it can swim to the top if the water speed is \(3.00 \mathrm{~m} / \mathrm{s}\). If water has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) as it passes over a ledge, (a) how far below the ledge will the water be moving with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) ? (Note that water undergoes projectile motion once it leaves the ledge.) (b) If the salmon is able to jump vertically upward from the base of the fall, what is the maximum height of waterfall that the salmon can clear?

A figure skater glides along a circular path of radius \(5.00 \mathrm{~m}\). If she coasts around one half of the circle, find (a) the magnitude of the displacement vector and (b) what distance she skated. (c) What is the magnitude of the displacement if she skates all the way around the circle?
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