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Chapter 3: Problem 67

A basketball player is 15 ft horizontally from the center of the basket, which is \(10 \mathrm{ft}\) off the ground. At what angle should the player aim the ball from a height of \(8.2 \mathrm{ft}\) with a speed of \(26 \mathrm{ft} / \mathrm{s} ?\)

Short Answer

Expert verified
The launch angle should be calculated using the equation \(\theta = \arctan(\frac{y}{x} + \frac{g x}{2 v^2})\). Substituting known quantities, the exact angle can be obtained.

Step by step solution

01

Identify given quantities

First, identify and write down all the known quantities. The horizontal distance to the basket (x) is 15ft, the height difference to the basket (y) is \(10ft - 8.2ft = 1.8ft\), and the launch speed (v) is 26ft/s.
02

Formulated equations

The motion of the basketball can be split into horizontal and vertical components. For horizontal motion, the equation is \(x = v \cdot t \cdot \cos(\theta)\). For vertical motion, the equation is \(y = v \cdot t \cdot \sin(\theta) - \frac{1}{2}gt^2\), where \(g = 32.2\) \(ft/s^2\) is the acceleration due to gravity and \(\theta\) is the launching angle. Notice that time (t) is the same for both motions.
03

Eliminate t

Eliminate t by dividing the y equation by the x equation, yielding \( \tan(\theta) = \frac{y}{x} + \frac{g x}{2 v^2}\)
04

Solve for the launching angle

Substitute the known quantities into the equation to obtain the launching angle \(\theta = \arctan(\frac{y}{x} + \frac{g x}{2 v^2})\)

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