acceleration due to gravity.docx - Lily Mills Acceleration Due to Gravity Practical 10G Aim To use three methods to measure the acceleration due to

acceleration due to gravity.docx - Lily Mills Acceleration...

This preview shows page 1 - 2 out of 2 pages.

Lily Mills Acceleration Due to Gravity Practical 10G Aim: To use three methods to measure the acceleration due to gravity. Method: Prac 1: Freefall: Drop a tennis ball off the balcony and measure the height through which it fell and the time it takes to reach the ground. Do this three times and take the average time. Prac 2: Ticker tape: Sticky tape a mass to the end of a piece of ticker tape and use a ticker tape timer to obtain three tapes of free fall. Prac 3: Simple Pendulum: Measure the time it takes for a simple pendulum to make ten complete oscillations (i.e. back to where it started is a complete oscillation) and the length of the pendulum. (Note small oscillations and a long length give the best results). Results: Prac 1: Drop Time (s) 1 1.3 2 1.7 3 1.53 Distance from balcony to ground was 7.6 metres. The average time it took for a tennis ball to reach the ground was 1.53 seconds.
Image of page 1

Subscribe to view the full document.

Image of page 2

Unformatted text preview: S= distance= 7.6 metres A= acceleration= T= time taken=1.53 seconds U= initial velocity= 0 Acceleration due to gravity: s = 1 2 at 2 + ut s − ut = 1 2 at 2 1 2 at 2 = s − ut at 2 = 2 ( s − ut ) a = 2 ( s − ut ) t 2 a = 2 ( 7.6 − ( × 1.53 ) ) 1.53 2 2 decimal places a = 6.49 ¿ ) metres per second Prac 2: a= (final/0.02-initial/0.02)/(number of dots between final and initial x 0.02) ms-2 final velocity: 10 initial velocity: 6 Lily Mills Acceleration Due to Gravity Practical 10G a = final velocity 0.02 − initialvelocity 0.02 ( final velocity − inital velocity ) × 0.02 a = 10 0.02 − 6 0.02 ( 10 − 6 ) × 0.02 a = 20.24 metres per second Prac 3: L=length= 30 cm T=period= 15.15 seconds G= G L ¿ ¿ ¿ T = 2 π √ ¿ T 2 π = √ L G T 2 4 π 2 = L G G = 4 π L 2 T G = 4 × π× 30 2 15.15 G = 746.52 ( 2 decimal places )...
View Full Document

  • Two '17
  • Britt

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

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes