This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework # 3 help 4.11 Model: The ball is treated as a particle and the effect of air resistance is ignored. Visualize: Solve: Using ( ) 1 1 x x x v t t = + , ( )( ) 1 50 m 0 m 25 m/s 0 s t = + m 1 2.0 s t ⇒ = Now, using ( ) ( ) 2 1 1 1 1 2 y y y y v t t a t t = + + , ( ) ( ) 2 2 1 1 2 0 m 0 m 9.8 m/s 2.0 s 0 s y = + + 19.6 m =  Assess: The minus sign with y 1 indicates that the ball’s displacement is in the negative y direction or downward. A magnitude of 19.6 m for the height is reasonable. 4.45. Model: The particle model for the ball and the constantacceleration equations of motion are assumed. Visualize: Solve: (a) Using ( ) ( ) 2 1 1 1 1 2 y y y y v t t a t t = + + , ( ) ( ) ( ) ( ) 2 2 1 2 0 m 30 m/s sin60 4 s 0 s 9.8 m/s 4 s 0 s h = + ° + = 25.5 m (b) Using ( ) ( ) 2 2 top top 2 y y y v v a y y = + , ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 top top 2 30 m/s sin60 sin 0 m /s sin 2 34.4 m 2 g 2 9.8 m/s v v g y y θ θ ° = + ⇒ = = = (c) The x and y components are...
View
Full
Document
This note was uploaded on 04/18/2010 for the course PHYS ? taught by Professor Zhou during the Spring '10 term at Georgia State University, Atlanta.
 Spring '10
 ZHOU

Click to edit the document details