Lecture 7.docx

# Lecture 7.docx - Projectile review Object of interest its...

• Notes
• 6

This preview shows 1 out of 3 pages.

Projectile review Object of interest, its only influenced by gravity and air resistance o Often, we neglect air resistance There are certain things that affect the outcome of the trajectory (path it takes) o angle of the trajectory and the trajectory speed magnitude and direction of the velocity at take off o relative height of projection whether they are at the same height or not Important to split the velocity in to the vertical and horizontal at every point in time o Vertical velocity gets smaller till we get to the top which it reaches 0 then get larger again going down Problem 7 (pg. 350) A soccer ball is kicked with an initial horizontal speed of 5 m/s and an initial vertical speed of 3 m/s. Assuming that projection and landing heights are the same and neglecting air resistance: o a) Identify the ball’s horizontal speed 0.5 s into its flight. 5 m/s as long as it hasn’t landed yet, the horizontal velocity is constant o b) Identify the ball’s horizontal speed midway through its flight. 5 m/s all the way through the flight it is constant o c)Identify the ball’s horizontal speed immediately before contact with the ground. 5 m/s still constant o d) Identify the ball’s vertical speed immediately at apex of the flight. 0 m/s at that top that is the one place we know the velocity of every projectile flight o e) Identify the ball’s vertical speed midway through the flight. 0 m/s because hallway through is the apex apex is only halfway through if take-off and landing are the same height o f) Identify the ball’s vertical speed immediately before contact with the ground. 3 m/s (down) vertical velocity is -3 m/s or 3 m/s down Additional Problem 10 (page 351) An archery arrow is shot with a speed of 45 m/s at an angle of 10°.

Subscribe to view the full document.

How far horizontally can the arrow travel before hitting a target the same height from which it was released? o Known: Projection speed: 45 m/s Projection angle: 10° Relative projection height: 0 m o Want: Range (horizontal distance) (d) Step 2: Draw sketch Step 3: Find horizontal and vertical components of take-off velocity o We need to split it into the 2 components V (vertical) = 45(sin10) = 7.81 m/s V (horizontal) = 45(cos10) = 44.32 m/s o these are the times where we stop and go does this make sense 45 is the hypotenuse always the largest side bigger than the other 2 so this is right
You've reached the end of this preview.

{[ 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