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Chapter 3: Problem 58

A \(2.00-\mathrm{m}\) -tall basketball player is standing on the floor \(10.0 \mathrm{~m}\) from the basket, as in Figure \(\mathrm{P} 3.58\). 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
The initial speed at which the player must throw the basketball to make the shot without hitting the backboard is approximately 11.2 m/s.

Step by step solution

01

Determine the Variables from the Problem

First, let's define the initial speed of the ball(v), the height of the ball above the ground(h), the distance from the basket(d), and the launch angle of the ball(a). \nGiven: \na = \(40.0^{\circ}\) \nd = 10.0m \nh = 3.05m - 2.00m = 1.05m
02

Determine the Vertical Component of Initial Velocity

The vertical and horizontal motions of the ball are independent of each other. Thus, the vertical motion can be solved as a one-dimensional kinematics problem.\nThe equation of motion we use involves the final position (y), initial position (y0), initial velocity (v0y), acceleration (a), and time (t).\nFinally, we use the equation \(y = y0 + v0yt + 0.5at^2\). We know y = 1.05m, y0 = 0, a = -9.8m/s^2, and y is upward therefore negative. Rearranging for v0y we have \[v0y = [(y - y0 - 0.5at^2)] / t\] where t = d / v0x.
03

Determine the Horizontal Component of Initial Velocity

Now, to find the horizontal velocity, we break down the resultant initial velocity into its components along the horizontal and vertical directions. The horizontal velocity for projectile motion remains constant all through the path.\nThe horizontal component of velocity can be computed with the formula for the x component of velocity in two dimensions, \[v0x= v0 cos(a)\]. But v0 is what we have left to find, so first we have to rearrange in terms of v0: \[v0 = v0x / cos(a)\]. But v0x is not known directly, it will be expressed in terms of the final vertical velocity: \[v0 = (d / t) / cos(a)\]. Where cos(40 degrees) = 0.77.
04

Substituting and Solving

In order to proceed, we must express t in terms of known values.\nSetting Step 2 equal to Step 3 to find an expression for v0 gives us a single equation with v0 as the unknown to be solved. Substituting the expressions for\(v0x\) and \(v0y\) from steps 2 and 3 we get:\n\[((y - y0 - 0.5at^2) / d ) cos(a)] = [(d / t) / cos(a)]\], which simplifies to: \n\[v0 = sqrt[-2ad + 2g(y0 - y) / cos^2 (a)]\]. Inserting the known values and calculating gives: \[v0 = sqrt[-2*10m*9.8m/s^2 + 2*9.8m/s^2 * 1.05m / (0.77)^2] = 11.2 m/s\].

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. When analyzing the motion of a projectile, like a basketball shot towards a hoop, we consider both horizontal and vertical components.

In the context of projectile motion:
  • The vertical motion is affected by gravity. This means that the vertical component of velocity changes over time.
  • The horizontal motion assumes there is no air resistance, meaning the horizontal velocity remains constant.
By breaking down the motion into these components, we can apply the basic equations of kinematics separately.

These equations help us determine various properties like time of flight, maximum height, and the range of the projectile. In our basketball problem, using kinematics allows us to find out the necessary conditions, such as the initial speed required for the ball to reach the basket.
Initial Velocity
Understanding initial velocity is crucial in projectile motion as it determines both how far and how high the object will go. Initial velocity can be broken into two components:
  • Horizontal component (\( v_{0x} \)): Determines how far the projectile will travel horizontally. It is calculated as \( v_{0x} = v_0 \cos(\theta) \), where \( v_0 \) is the initial speed and \( \theta \) is the launch angle.
  • Vertical component (\( v_{0y} \)): Determines how high the projectile will rise and is calculated as \( v_{0y} = v_0 \sin(\theta) \).


In the basketball scenario, calculating initial velocity requires knowing the vertical and horizontal distances and rearranging the kinematic equations. The vertical motion, affected by gravity, influences the ball's ascent and descent, while the horizontal component ensures it reaches the basket at the desired location. Solving for initial velocity involves manipulating these components mathematically to meet the criteria set by the problem conditions, such as angle and distance.
Projectile Range
Projectile range denotes the horizontal distance traveled by a projectile. To calculate this, we need a clear understanding of both the launch angle and initial velocity. In idealized conditions (without air resistance), the formula for projectile range (\( R \)) is:

\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]

where:
  • \( v_0 \) is the initial speed,
  • \( \theta \) is the launch angle,
  • \( g \) is the acceleration due to gravity (\( 9.8 \text{ m/s}^2 \) on Earth).


However, the situation with the basketball is slightly different as we need the ball to pass through a point at a specific height, making it a bit more complex than reaching maximum range. Therefore, the goal is to adjust initial speed and angle to meet the basket's height rather than simply maximizing the range. By understanding the relationship between these factors, we can calculate the precise initial velocity needed to ensure the ball passes through the hoop at the right distance, further integrating concepts from initial velocity and kinematics.

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Most popular questions from this chapter

A small map shows Atlanta to be 730 miles in a direction \(5^{\circ}\) north of east from Dallas. The same map shows that Chicago is 560 miles in a dircction \(21^{\text {n }}\) west of north from Atlanta. Assume a flat Earth and use the given information to find the displacement from Dallas to Chicago.

A river flows due east at \(1.50 \mathrm{~m} / \mathrm{s}\). A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) due north relative to the water. (a) What is the velocity of the boat relative to the shore? (b) If the river is \(300 \mathrm{~m}\) wide, how far downstream has the boat moved by the time it reaches the north shore?

A home run is hit in such a way that the baseball just clears a wall \(21 \mathrm{~m}\) high, located \(130 \mathrm{~m}\) from home plate. The ball is hit at an angle of \(35^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall. and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of \(1.0 \mathrm{~m}\) above the ground.)

A chinook salmon has a maximum underwater speed of \(3.58 \mathrm{~m} / \mathrm{s}\), but it can jump out of water with a spced 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, how far below the ledge will the water be moving with a speed of \(9.00 \mathrm{~m} / \mathrm{s}^{2}\) (Note that water undergoes projectile motion once it leaves the ledge.) 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 rocket is launched at an angle of \(53.0^{\circ}\) above the horizontal with an initial speed of \(100 \mathrm{~m} / \mathrm{s}\). The rocket moves for \(3.00 \mathrm{~s}\) along its initial line of motion with an acceleration of \(30.0 \mathrm{~m} / \mathrm{s}^{2}\). At this time, its engines fail and the rocket proceeds to move as a projectile. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and \((c)\) its horizontal range.
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