Shoot to Win! and Parabolic Motion

Shoot to Win Lab

Hit the textbook- 100 points
Miss the textbook- do a lab report

No one wanted to have to do a lab report. Everyone wanted a 100! So we needed to solve this problem:

Mr. Gray will shoot the gun at a horizontal angle. Given that the gun shoots the ball at a constant velocity, predict where the ball will land when the gun is shot at a 52 degree angle.

Part One: The Horizontal Shot

What do we know?
1. We know that the initial velocity in the y direction is 0.
2. We know the acceleration in the y direction is -g (-9.8m/s)
3. We know the acceleration in the x direction is 0.

What do we need to find from the initial shot (a straight horizontal)?
1. How far the ball goes in X and Y components (deltaY and deltaX)
2. Time it takes for the ball to land (t)

Mr. Gray shot the gun at the horizontal. We measured how far away from the table it landed (1.74m) and the height of the table (1.055m).

Now we have this for the y components:

Let's solve for time.
deltaY = Vi(t) + 1/2at^2
-1.055 = 0 + 1/2(-9.8)t^2
t = 0.46s

Then we filled in the x component side of the chart, as well as the time:

We can determine the velocity the ball shoots out of the gun from the x components.
deltaX = Vi(t) +1/2at^2
1.74 = Vi(0.46) + 0
Vi = 3.7 m/s

Now we know the velocity that the gun shoots the ball at! 3.7m/s

Part Two: The Angled Shot

Using this velocity, we can split the velocity into components using the 52 degree angle. The y component becomes 3.7sin52 = 2.92 and the x component becomes 3.7cos52 = 2.28
We also remeasured the height of the gun because it changed as the angle changed. The height was 1.135m. We still know that acceleration for y is -9.8 and 0 for x. So let's fill out a table:

What do we want to solve?
1. We want to solve for time in the y component- which is also time in the x component.
2. By solving for time, we can then solve for deltaX, which determines where the textbook should be placed.

Solve for t.
deltaY = Vi(t) + 1/2at^2
-1.135 = 2.92t +1/29.8)t^2
t = 0.868s

Update the chart:

Finally, we can solve for deltaX!
deltaX = Vi(t) +1/2at^2
deltaX = 2.28(0.868) + 0
deltaX = 1.98m

We placed the center of the textbook (to account for errors) at 1.98m away from the table. SUCCESS!