Experiment Results: Projectile Motion and Velocity Analysis

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SUMMARY

The experiment analyzed projectile motion and velocity of a ball rolling off a table with a height of 86.4 cm. The horizontal velocities calculated were 91 cm/s for trial 1 and 83 cm/s for trials 2 and 3. The vertical component of the ball's velocity upon leaving the table is confirmed to be 0. The time to reach the floor was calculated using the formula d = v_{y_{0}}t + 0.5gt², with the correct height of 86.4 cm and g as 980 cm/s². It is essential to maintain consistent units throughout the calculations.

PREREQUISITES
  • Understanding of basic kinematics and projectile motion principles
  • Familiarity with the equations of motion, specifically d = v_{y_{0}}t + 0.5gt²
  • Knowledge of unit conversion, particularly between centimeters and meters
  • Experience with calculating average velocity using v = d/t
NEXT STEPS
  • Learn about the effects of air resistance on projectile motion
  • Study the derivation and application of the equations of motion in different contexts
  • Explore advanced projectile motion simulations using software like PhET Interactive Simulations
  • Investigate the impact of varying launch angles on horizontal distance traveled
USEFUL FOR

Students in physics, educators teaching projectile motion, and anyone conducting experiments related to kinematics and motion analysis.

courtrigrad
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I did an experiment which involved a few trials. Here are the results:

distance rolled on table: 100 cm (all three trials) .
height of table: 86.4 cm
time on table: 1.1 s (trial 1), 1.2 s (trial 2), 1.3 s (trial 3) .
horizontal distance traveled in air: 36.1 cm (trial 1), 34.8 cm (trial 2), 34.8 cm (trial 3) .

Ok so (1) What is the velocity the ball rolled with on the table? So [tex]v = \frac{d}{t}[/tex]. I got 91 cm/s (trial 1), 83 cm/s (trial 2), 83 cm/s (trial 3).

(2) The vertical component of the balls velocity when it leaves the edge of the table would be 0. Is this correct?

(3) Calculate the time that it takes the ball to reach the floor once it has left the table. Ok so I would use [tex]d = v_{y}_{0}t + \frac{1}{2}gt^{2}[/tex]. [tex]v_{y}_{0} = 0, d = 86.4 cm[/tex]. Are these the correct values for the variables?

(4) Find the horizontal distance that the ball should travel from the time it left the table until it hit the floor for the three trials. So would I use [tex]x = x_{0} +v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}[/tex]? So would I use the 91 cm/s, 83 m/s, and 83 m/s, for the velocities, and 0.42 seconds for the time?

Thanks
 
Last edited:
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You are mixing up units for part (3), make sure you convert to one set of units.
 

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