Projectile Motion lab

In order to demonstrate rpoject motion, we threw a basketball in the air. We videotaped this motion using video physics, and we plotted points along the flight path of the basketball. We were then able to creat graphs with which we examined the projectile motion of the basketball. 

Graph of x velocity 

Mh linear fit equation was y= -1.740 x+ 3.424. Since the slope is relatively close to 0, it is a horizontal line. Since the slope is 0, the acceleration is 0 in the x direction. The object is moving to the left because the velocity is negative. There is no horizontal force, so it is traveling at a constant speed horizontally. The equation is Vx(t)= Vxi. 

Graph of y velocity: 

My linear fit equations was y= -11.336x + 5.245. When the velocity is postive, it is slowing down and moving upward. At Vy= 0, it turns around, and speeds up moving downward due to the negative velocity. The velocity of y is constantly changing with time. The slope is the acceleration due to velocity. 

The equation is Vy(t)= -10m/s2t + Vyi. These graphs always have a slope of -10m/s2 and are never curved. 

Graph of x:

My linear fit equation was y= 2.746x + .089. The slope of x vs t is Vx. The object is moving in the t direction and rightward. The equation is x(t)= Vxt+ Xi. These graphs are never curved because velocity is constant. 

Graph of y:

The equation of my linear fit line was y= 1.139x + .587. The graphs for y are always curved as it demonstrates the flight path. The equation is y(t)= (-10m/s2) t2 + Vyit +Yi.