110_Physics ProblemsTechnical Physics

# 110_Physics ProblemsTechnical Physics - Chapter 4 P4.67 (a)...

This preview shows page 1. Sign up to view the full content.

Chapter 4 111 P4.67 (a) xv t xi = , yv t g t yi =+ 1 2 2 dt cos . . cos . 50 0 10 0 15 0 °= ° af and −° = ° + t sin . . sin . . 50 0 10 0 15 0 1 2 980 2 a f . Solving, d = 43 2 . m and t = 288 . s . (b) Since a x = 0, FIG. P4.67 vv vva t xf xi yf yi y == ° = =+= ° = 10 0 15 0 9 66 10 0 15 0 9 80 2 88 25.6 .. m s ms cos . .s i n . . . . Air resistance would decrease the values of the range and maximum height. As an airfoil, he can get some lift and increase his distance. *P4.68 For one electron, we have t iy = , Dv t a t a t ix x x 1 2 1 2 22 , yf yi = , and v v at at xf xi x x =+≅ . The angle its direction makes with the x -axis is given by θ === = −−− tan tan tan tan 111 2 1 2 v v v at vt y D yf xf yi x yi x . FIG. P4.68 Thus the horizontal distance from the aperture to the virtual source is 2 D . The source is at coordinate xD =− . *P4.69 (a) The ice chest floats downstream 2 km in time t , so that 2 km = w . The upstream motion of the boat is described by dvv w ( )15 min. The downstream motion is described by
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online