CHAPTER 2 - 12 - 0=87.6m/s+(9.80m/s )(t 27.4s),whichgivest...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
v  = 0 y  = 0 H a + v 0 0 = 87.6 m/s + (– 9.80 m/s 2 )( t 2  – 27.4 s), which gives  t 2  =       36 s . ( e ) We consider the motion after the rocket runs out of fuel: v 3 2  =  v 1 2  + 2(–  g )( y 3  –  y 1 );    v 3 2  = (87.6 m/s) 2  + 2(– 9.80 m/s 2 )(0 – 1200 m),  which gives  v 3  = – 177 m/s =        – 1.8 × 10 2  m/s ( f ) We find the time from v 3  =  v 1  + (–  g )( t 3  –  t 1 )    – 177 m/s = 87.6 m/s + (– 9.80 m/s 2 )( t 3  – 27.4 s), which gives  t 3  =       54 s . 61. We use a coordinate system with the origin at the top of the window and down positive.  We can find the velocity at the top of the window from the motion  past the window:   y  =  y 0  +  v 0 t  +  ! at 2 ;    2.2 m = 0 +  v 0 (0.30 s) +  ! (9.80 m/s 2 )(0.30 s) 2 , which gives   v 0  = 5.86 m/s. For the motion from the release point to the top of the window, we have
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

Page1 / 2

CHAPTER 2 - 12 - 0=87.6m/s+(9.80m/s )(t 27.4s),whichgivest...

This preview shows document pages 1 - 2. Sign up to view the full document.

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