Answered: Physics Question

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Expert Solution

Step 1

Given

The height of the player is h_p = 2 m.

The basket height is h_b = 3.05 m.

The horizontal distance between the player and basket is L = 12 m.

The angle with the horizontal is θ = 38^o.

The horizontal component of the velocity is,

$\begin{array}{rcl} v_{x} & = & v \cos θ \\ = & v \cos 38 \\ = & 0.788 v \end{array}$

The vertical component of the velocity is,

$\begin{array}{rcl} v_{y} & = & v \sin θ \\ = & v \sin 38 \\ = & 0.616 v \end{array}$

Step 2

The horizontal distance covered by the ball due to horizontal velocity is cacluted as,

$\begin{array}{rcl} L & = & v_{x} \times t \\ 12 m & = & 0.788 v \times t \\ v & = & \frac{12 m}{0.788 \times t} \\ v & = & \frac{15.23 m}{t} . . . . (i) \end{array}$

The vertical distance covered by the ball due to vertical velocity is cacluted as,

$\begin{array}{rcl} ∆ h & = & v_{y} t + \frac{1}{2} (- g) t^{2} \\ h_{b} - h_{p} & = & 0.616 v \times t - \frac{1}{2} \times 9.8 m / s^{2} \times t^{2} \\ 3.05 m - 2 m & = & 0.616 v \times t - 4.9 m / s^{2} \times t^{2} \\ 1.05 m & = & 0.616 v \times t - 4.9 m / s^{2} \times t^{2} \end{array}$