Velocity calculation 151 0478 v x 316 m s calculated

  • No School
  • AA 1
  • 5

This preview shows page 4 - 5 out of 5 pages.

Velocity calculation 1.51 0.478 = v 0 x = 3.16 m / s Calculated values: Measurement Value Time (sec.) 0.478 Velocity (m/s) 3.16 Part B: Measurement Value vertical distance (m) 1.131 Velocity (m/s) horizontal 2.72 Velocity vertical (m/s 1.57 Theoretical horizontal range (m) 1.82 Experimental Horizontal Range (m) 1.8268 Calculations B: Vector Velocity: 3.14cos ( 30 ) = x = 2.72 3.14sin ( 1.57 ) = y = 1.57 Time elapsed: 1.131 = 1.57 t ± 4.9 t 2
Image of page 4
5 EVALUATION OF PROJECTILE MOTION t = 0.67 Distance traveled: 2.72 × 0.67 = ∆ X = 1.8224 m. Percent Error: | 1.8268 1.8224 | 1.8224 × 100 = 0.24 % error Discussion The results of this experiment helped to shine light on the factors that affect projectile motion. Though the horizontal vector velocity of the second part of the experiment is slower than the first part, the horizontal distance is significantly larger in the second part. The difference in results of this experiment can be attributed the launch angle of the projectile, causing a different trajectory. An extension to this experiment could be calculating and testing angles of trajectory in order to identify the best angle for the most horizontal distance. This could also be tested with projectiles of differing weights to determine how mass effects distance. The minor error in the final distance could have been due to the lack of recognition to air resistance in the calculation, which was not a hindrance to the final result. Overall, the experiment was successful and provided satisfactory results.
Image of page 5

You've reached the end of your free preview.

Want to read all 5 pages?

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors