For this project, we would throw balls of various sizes and weights from 1 yard in the air and time how long they were in the air. Now we are finding the graph of the ball's path.

The general equation for finding the path of a projectile is . We can fill in parts of this equation to form a quadratic function. Since I threw two balls, I will have two equations. When you fill in the first equation you get h=-16(1.5^2) + Vo(1.5) + 3. The second equation is -16(1.6^2) + Vo(1.6) + 3.

To find Vo, we will use the above equation, and put in 0 for the final height. So 0=-16(1.5^2) + Vo(1.5) + 3 and 0=-16(1.6^2) + Vo(1.6) + 3

V0=21.31 Vo=24.41

We can fill these back into the equation:

h=-16t^2 + 21.3t + 3 and h=-16t^2 +24.4t + 3

To find what time the balls reached their maximum height, we can use the equation

aos=(-b/2a). The axis of symmetry (aos) tells us the time the projectile reaches its maximum

height. So the time the ball reaches its maximum height=(-21.3/-32)=0.7 and (-24.4/-32)=0.8

You can now use the axis of symmetry to find what the y-coordinate, or highest height, is,

by plugging in the aos into the equation as the x(t, in this case)-values. H will equal the maximum

height because the time is at the time the ball was at its maximum height:

h=-16(0.67^2) + 21.31(0.67) + 3=10.1 and h=-16(0.76^2) +24.41(0.76) + 3=12.3

The maximum heights are 10.1 feet and 12.3 feet.