Video Analysis Blog Post

1. In this experiment, we learned the basics of shooting and analyzing a video containing a ball with upward and forward motion on PASCO Capstone. We began filming our video by putting a meter stick on the wall so the system would have a basis for measurement, and later analyzed it in the system. The video is below.

 

2. 

a. Below is a picture of the markings that represent the motion of the ball after it was passed, or the the ball’s ascending and descending motion. The marks were used to create the velocity-time and position-time graphs that will be displayed later.

b. This position versus time graph represents the position of the ball in both the horizontal and vertical directions. The curved line represents the motion in the vertical direction. The ball moves upward at a constant velocity, passes through a velocity of 0m/s at the top of its path, then falls downward at a constant velocity. The straight diagonal line represents movement in the horizontal direction. It is straight because the ball was traveling at constant velocity moving forward in the horizontal direction.

c. This is the velocity versus time graph which also indicates motion in the horizontal and vertical directions. The blue line represents motion in the vertical direction. The line with a negative slope starts above the x axis which represents the ball slowing down as it reached the top of its path, and as it crosses the x axis and continues downward, it represents the ball speeding back up as it travels downwards into Dohee’s hands. 

3. Analysis

a. ACCELERATION IN THE Y (VERTICAL) DIRECTION: 9.88 m/s^2- This was found by simply finding the slope of the velocity time graph, and I know this is accurate because the acceleration for any falling object is always 9.8m/s^2 because it it the universal gravitational constant,  but for the sake of this class we round it up to 10. 9.88 is pretty accurate, and it is a negative because the slope of the line on the v-t graph is negative.

b. ACCELERATION IN THE X (HORIZONTAL) DIRECTION: 0 m/s^2- I know this to be true because the ball is traveling at a constant velocity (as indicated by the position time graph) because the forces are balanced in the horizontal direction, therefore, nothing is acting on it to cause acceleration

c. INITIAL VELOCITY IN X DIRECTION: -.87 m/s- I know this to be true because this is the y intercept for the orange line on the v-t graph which represents initial velocity in the horizontal direction

d. INITIAL VELOCITY IN Y DIRECTION: -3.24 m/s- I know this to be true because this is the y intercept for the blue line on the v-t graph. The y intercept represent the initial velocity in the vertical direction

e.VELOCITY AT THE TOP OF PATH IN HORIZONTAL DIRECTION: constant non zero velocity- I know this to be true because the ball is always moving at some given speed in the horizontal direction, but the acceleration at the top of the path in the horizontal direction is 0m/s^2, because once again, the forces are balanced in the horizontal direction 

f. VELOCITY AT THE TOP OF PATH IN VERTICAL DIRECTION: 0m/s- I know this is true because the ball temporarily stops at       the top of its path before it changes direction,but the acceleration is -10m/s^2, because that is always the accretion for anything that is falling

g. FINAL VELOCITY IN THE X DIRECTION: -.25 m/s- This is the final point on the v-t graph for the orange line

h.  FINAL VELOCITY IN THE Y DIRECTION: -1.74 m/s- This is the final point on the v-t graph for the blue line

i. HOW HIGH THE BALL GOT: 1.83 m- I found this by locating the highest point on the position time graph along the arc, which is movement in the vertical direction. This also can be found using the displacement formula, change in x= 1/2(a)t^2 +(vi)t. To solve for displacement, you would simply plug in the values from the graph including the acceleration, time, and initial velocity, and arrive at the same answer

j. HOW FAR THE BALL WENT: To find displacement, I used the formula xfinal-xinitial. According to the graph, the ball’s final position was .51 and the initial was .48, and according to the formula the ball traveled .03meters 

k. HOW MUCH TIME THE BALL TOOK TO GET TO THE TOP OF ITS PATH: 1.89 seconds- this is the x coordinate that matches with the highest point on the graph of 1.83m

l. HOW MUCH TIME THE BALL WAS IN THE AIR: 2.453 seconds- this is the final time according to the position time graph 

4. Conclusions

a. From this experiment, I can have learned that the acceleration in the vertical and horizontal directions are different despite my assumption that they will be related in some way. 

b.The vertical versus horizontal formulas for finding velocity are also different and must not be confused for one another or else the wrong answer will be found. In the horizontal direction, the formula to solve for velocity is v= change in position/change in time, but in the vertical direction, you have to manipulate the formula acceleration=change in velocity/ change in time to solve for velocity. 

c. The vertical versus horizontal formulas to solve for displacement are different as well. For the horizontal direction, you would use the formula displacement= final position-initial position, but for the vertical direction, you can either solve for the area under the line, or use the formula displacement=1/2(a)t^2+(vi)(t).

d. The vertical formula to solve for height is also 1/2(a)t^2+(vi)t

e. Staying true to the trend, the vertical versus horizontal velocities at the top of the ball’s path differ, as the velocity is 0m/s in the vertical direction because the ball goes through a brief period of rest, and it is at a constant non zero velocity in the horizontal direction because it is constantly moving forward.

To conclude, I have learned how distinct the horizontal and vertical directions are when it comes to motion and displacement, and there really isn’t any overlap when it comes to calculating certain values.