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Q67E

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Found in: Page 125

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# An ice hockey player is moving at m/s when he hits the puck toward the goal. The speed of the puck relative to the player is m/s. The line between the center of the goal and the player makes a angle relative to his path, as shown in Figure. What angle must the puck's velocity make relative to the player (in his frame of reference) to hit the center of the goal?

The angle that the puck’s velocity must make relative to the player is $73.{98}^{°}$ .

See the step by step solution

## Step 1: Definition of velocity

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

The speed of the puck relative to the player is 29 m/s.Let’s focus on the triangle in the question.

## Step 2: The angle that the puck’s velocity must make relative to the player

The angle that the puck’s velocity must make relative to the player can be calculated as:

$\begin{array}{rcl}\mathrm{sin}\theta & =& \frac{{V}_{player}}{{V}_{puck}}\\ \mathrm{sin}\theta & =& \frac{8}{29}\\ \mathrm{sin}\theta & =& 0.28\\ \theta & =& {\mathrm{sin}}^{-1}\left(0.28\right)\\ \theta & =& 16.{01}^{°}\end{array}$

The angle that the puck’s velocity must make relative to the player is:

$\beta ={90}^{°}-\theta \phantom{\rule{0ex}{0ex}}\beta ={90}^{°}-16.{01}^{°}\phantom{\rule{0ex}{0ex}}\beta =73.{98}^{°}$

The angle that the puck’s velocity must make relative to the player is $73.{98}^{°}$ .