{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

problem02_86 solution

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

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

2.86: a) The helicopter accelerates from rest for 10.0 s at a constant 2 s m 5.0 . It thus reaches an upward velocity of s m 50.0 s) 0 . 10 )( s m 0 . 5 ( 2 0 = = + = t a v v y y y and a height of m 250 s) 0 . 10 )( s m 0 . 5 ( 2 2 2 1 2 2 1 = = = t a y y at the moment the engine is shut off. To find the helicopter's maximum height use ) ( 2 0 2 0 2 y y a v v y y y - + = Taking m 250 0 = y , where the engine shut off, and since 0 2 = y v at the maximum height: m 378 ) s m 8 . 9 ( 2 s) m 0 . 50 ( m 250 2 2 2 max 2 0 0 max = - - = - = - y g v y y y or 380 m to the given precision. b) The time for the helicopter to crash from the height of 250 m where Powers stepped out and the engine shut off can be found from: 0 ) s m 8 . 9 ( s) m 0 . 50 ( m 250 2 1 2 2 2 1 2 0 0 = - + + = + + = t t t a t v y y y y where we now take the ground as 0 = y . The quadratic formula gives solutions of s 67 . 3 = t and 13.88 s, of which the first is physically impossible in this situation. Powers'
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: position 7.0 seconds after the engine shutoff is given by: m 9 . 359 ) s . 49 )( s m 8 . 9 ( 2 1 s) . 7 )( s m . 50 ( m 250 2 2 =-+ + = y at which time his velocity is s m 6 . 18 ) s . 7 )( s m 80 . 9 ( s m . 50 2-=-+ = + = gt v v y y Powers thus has s 88 . 6 . 7 88 . 13 =-more time to fall before the helicopter crashes, at his constant downward acceleration of 2 s m 2.0 . His position at crash time is thus: 2 2 1 t a t v y y y y + + = 2 2 ) s 88 . 6 )( s m . 2 ( 2 1 s) 88 . 6 )( s m 6 . 18 ( m 9 . 359-+-+ = m 6 . 184 = or 180 m to the given precision....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern