Hit-The-Target


 * Lab Title:** Hit-The-Target!!


 * Primary Authors:** Kelsey


 * Contributing Authors:** Carson and Kerry


 * Methods:** While preforming the lab we wrote down the information that we were given, and then used the multiple equations (listed below) to help find the unkowns that we wanted. Then to test out our results we would place the catcher/board in the right place in order to see if we had done that lab correctly.

Part 1.


 * Goal:** To determine where to place a catcher for a horizontally laugnched projectile.


 * Data:** (medium speed)

Launch speed- Distance:10cm Time: 0.0204seconds Speed=Distance/Time (10)/(0.0204) = **4.901m/s** Launch height - Container height = total vertical displacement 1.17m - 0.13m = **1.04m**

Finding time for the vertical displacement 1.04 = 0*T + (.5)(-9.8)(T^2) T = **0.4607seconds**

Horizontal displacement = (4.901)(0.4607) + (.5)(0)(0.4607^2) Place Container **225.83meters** away

Part 2.


 * Goal:** To determine where to place a catcher for a projectile launched at an angle.


 * Data:** (Angle = 40 degrees)

When launched straight up the displacement was 0.76meters 0=Vo^2 + 2(-9.8)(0.76) Vo = **3.8595m/s** Launch height - Container height = total vertical displacement 1.11m - 0.13m = **0.98m**

Draw a triangle to aid in the next two steps:

Law of sines (3.8595)/(sin90) = Y/(sin40) Y = **2.4808m/s**

Law of sines (3.8595)/(sin90) = X/(sin60) X = **3.34m/s**

Finding time for the vertical displacement 0.98 = (2.4808)t + .5(-9.8)(T^2) T = **0.76703128seconds** Horizontal displacement = (3.34)(0.76703128) + (.5)(0)(0.76703128^2) Place container **2.5637meters** away

Part 3.


 * Goal:** To determine where on the vertical board a mini-launcher projectile launched at medium speed will hit.


 * Data:** (Angle = 56 degrees, Distance = 1.3m)

Triangle here

Law of sines (4.0089)/(sin90) = Y/(sin56) Y = **3.323603253m/s**

Draw a triangle to aid in the next two steps: Law of sines (4.0089)/(sin90) = X/(sin34) X = **2.2417987m/s** Finding time for the horizontal displacement 1.3 = (2.2417987)t + .5(0)(t^2) T = **0.57989seconds** Vertical displacement = (3.323603253)(0.57989) + .5(-9.8)(0.57989^2) Place the sheet with the 5 centered at **0.27958meters**


 * Conclusion: Overall our group did very well. We came very close to hitting our target and for one of them we even ended up hitting it! We discovered that to determine the distance a target needed to be due to angle and projectile motion, and we solved for all of our variables to hit the target. The factors that could make our ball not go into the target could have been anything from human error, a misfire or a change in angles due to tools that moved around.**