Assignment 3 Solutions (Fall 2008)

Assignment 3 Solutions (Fall 2008) - P207 - Fall 2008...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P207 - Fall 2008 Solutions Assignment 3 1 (a) Horizontal and vertical components of velocity are shown below. 10 20 30 40 50 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 v x [m/s] time [seconds]-5-4-3-2-1 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 v y [m/s] (b) The horizontal velocity 161km/hr = 161000/3600 = 44.7 m/s. The dis- tance to the plate is 18.2m. Then the time for the ball to get half way to the plate is t half = x half v x = 18 . 3 / 2 m 44 . 7 m / s = 0.205 s (c) The horizontal velocity does not change so the time to travel the second half is the same as the time to travel the first half . (d) The total time of 0.41 s is just twice the human reaction time of . 2 s. (d) y = v y t half- 1 2 gt 2 half . The initial vertical velocity v y = 0. Therefore, y =- 1 2 (10 m / s 2 )(0 . 205s) 2 =-0.210 m 1 (e) The change in vertical position over the second half of the horizontal distance is again. y = v y t half- 1 2 gt 2 half . But this time the initial vertical velocity is not zero. v y =- gt half . That is, v y is the vertical velocity gained during the first half of the trip. Then y = v y t half- 1 2 gt 2 half = (- gt half ) t half- 1 2 gt 2 half =- 3 2 gt 2 half =- 3 2 (10m / s 2 )(0 . 205m / s) 2 =-0.630 m (f) The change in vertical position is greater during the second half of the trip because the initial vertical velocity for the second half of the trip is non zero . 2. (a) The vertical component of velocity at launch is v y (0) = (100m / s) sin(30) = 50m / s. The vertical velocity vs vertical position is given by v 2 y = v 2 y (0)- 2 g ( y- y (0)) The initial vertical position is y (0) = 0. Solving for the height y when v y = 0, we get y = v 2 y (0) 2 g = (50m / s) 2 2(10m / s 2 ) = 125 m (b) We have that the vertical component of the velocity is v 2 y = v 2 y (0)- 2 gy . Therefore v y = ( v 2 y (0)- 2 gy ) 1 2 We know that v y (0) = 50 m/s, and g = 10m / s 2 . The maximum height is 125 m. The horizontal component of the velocity is (100 m / s) cos(30) = 86 . 6 m / s. So fraction of maximum height y [m] v x [m/s] v...
View Full Document

This note was uploaded on 10/26/2008 for the course PHYS 2207 taught by Professor Liepe, m during the Fall '07 term at Cornell University (Engineering School).

Page1 / 6

Assignment 3 Solutions (Fall 2008) - P207 - Fall 2008...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online