6lecture2a08

6lecture2a08 - Physics 2A Lecture 6: Jan 18 Vivek Sharma...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 2A Lecture 6: Jan 18 Vivek Sharma UCSD Physics
Background image of page 1

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

View Full DocumentRight Arrow Icon
QUIZ 2 SCORE seek help before its too late!
Background image of page 2
Equations Of Motion For Object under Constant Acceleration x0 x x vv a t =+ 0 2 x 0 x vt 1 at 2 x x + 22 x x 0 2 a ( x x ) 0x x 0 2 x t x + = ⎜⎟ ⎝⎠ Lets do some semi-soft & hard problems based on these ! Can replace x with y when things go horizontal to vertical
Background image of page 3

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

View Full DocumentRight Arrow Icon
Touchdown On The Moon A lunar lander is making its descent to moon base. The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0m above surface and has a downwa 2 moon rd speed of 0.8m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches surface. g =1.6m/s Apply constant acceleration equations to the motion of the lander Let downward be positive. Lander is in freefall a y =g moon 2 0y 0 y What we know: v =+0.8m/s, y-y =5.0m, a 1.6 / , no idea about t! ms = 22 y0 y 0 2 y y 0 Use v v 2 (y-y ) v = v +2a (y-y )= (0.8m/s) +2(1.6m/s )(5.0m) = 4.1m/s y a =+ y The same descent on Earth would have led to v =9.9m/s due to the stronger acceleration due to gravity g.
Background image of page 4
Coming to a lecture near you !
Background image of page 5

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

View Full DocumentRight Arrow Icon
Spiderman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/4 in the last 1.0s of his fall. What is the height h of the building? h +y -y 2 With this convention 9.8 / y am s =+ h/4 3h/4 Which equation to use? depends on what we know? 22 00 y y y0 y 0 2 y Use v Motion from roof to h/4 above ground 3/ 4 , 0,a =g So we get: v 0+ (3/ 4 ) v 2 3.834 m/s Spiderman's speed after he has v2 ( y - y ) fallen yy y h v ah h a ⇒− = = =⇒ == y for 3/4h is v 3.83 h = 0 2 y 0y In the next segment, / 4,v clearly we should use: y=y (1/ 2) ( 3.83 m/s, a =g, t=1s 3.83 m / 4) 4.90 . Now solve for ...but how ? y h vt at hm h h h −= = ++ ⇒= + Divide Spidey's motion in 2 segmen y = 0 y = 3/4 ts: and y = 3 h h /4h 2 2 2 2 4 if a 1 Write as 3.83 4 u0 2 Taking the positive root u=1 .90 0, solve for u (quadrati 273 c 6.52 e m q) 4 bb a c bu c u hu um a m −± = ⇒⇒ −− =
Background image of page 6
A Groovy Crash !
Background image of page 7

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

View Full DocumentRight Arrow Icon
2 Helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s . Austin Powers jumps on just as the helicopter lifts off the groun After the two men struggle for 10.0s, Powe d. rs s Assume that the chopper is in free fall after its engine is shut off, and ignore the effects of air-r h e uts off the engine and ste . . (a) what i ps out of s the max. the chopp height a er bo sist ve ence grou 2 Powers deploys a jet pack strapped on his back 7.0s after leaving chopper, and then has a constant downward acceleration with magnitude 2.0m/ nd reached by chopper.
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/08/2008 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.

Page1 / 29

6lecture2a08 - Physics 2A Lecture 6: Jan 18 Vivek Sharma...

This preview shows document pages 1 - 9. Sign up to view the full document.

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