problem02_82 solution

University Physics with Modern Physics with Mastering Physics (11th Edition)

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

View Full Document Right Arrow Icon
2.82: a)
Background image of page 1

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

View Full DocumentRight Arrow Icon
b) From the speed found in part (c), the maximum height is . m 635 ) s m 80 . 9 ( 2 ) s m 6 . 111 ( 2 2 = c) After the fuel is burned, the rocket has speed (40.0 2 s m )(2.50 s) = 100 m/s and has reached a height ( 2 1 )(40.0 2 s m )(2.50 s) 2 = 125 m. The speed of the rocket just before it hits the ground is then , s m 6 . 111 ) m 125 )( s m 80 . 9 ( 2 ) s m 100
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( ) ( 2 2 2 2 =--=--= y y g v v or s m 112 to three significant figures. d) The time from launch to the highest point is not the same as the time from the highest point back to the ground due to the upward acceleration of the engine in the first 2.5 s....
View Full Document

Page1 / 2

problem02_82 solution - ( ) ( 2 2 2 2 =--=--= y y g v v or...

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