Original Plan: v = 250 mph = 111.736m/s H = 2650 m Vx Ground Drop Zone R The plane is flying at a constant speed of 250 mph and is 2650 meters above the ground. v = Velocity of the plane v = 250mph which = 111.736 m/s h = height of the plane h = 2650 m 1 mile = 1.609 km 1.609 km x 250 miles = 402.25- - km mile hour hr Initial Horizontal velocity - 402.25 km/h - 111.736 m/s 5 X 16 = 111.736 m/s 402.25.The horizontal velocity stays constant when the payload is dropped because there is no horizontal component of force. The vertical velocity of the payload is 0 m/s when the payload is dropped and increases due to gravity. Initial vertical velocity = 0m/ s The object due to gravity is denoted by g (standard gravity) 4 2 8 = 9.80 m S Horizontal distance to the drop site can be calculated by r = v . / Using the equation of motion, we'll calculate the horizontal distance())2 - =2gh Initial Horizontal velocity = , = 111.7361 m/s Final Horizontal velocity = "! = V, (no horizontal force) Initial Vertical velocity = ", =0 ( v!) 2 - 03 = 2 . 9.8 . 2650 ( WS (2 = ( $1940 my -1) : -1 "! = \\51940 m =227.90 -1 S S = 227.90 = 0 + 9.8x 227.90 5 - 23.2553 9.8 Distance = vt = 111.7361m/s * 23.255s = 2598.281m The payload should be dropped at a distance of 2598.28m away from the drop zone. The velocity of the payload when it reaches the ground is denoted by: =V(111.7361) 3 + ( 227.90) = m/s = \\/2484.956 + 51938.41 m/ s = 253.817 m/s 1 So, the payload will hit the drop zone with a velocity of 253.817 mio What is the relationship between vertical and horizontal motion in kinematics equations? The horizontal velocity stays constant because there is no horizontal component of force. The vertical velocity of the payload is G m-'s when the payload is dropped and increases due to gravity. So= the horizontal and vertical motions in kinematics are related to each other throughout time. o How did you analyze the vertical motion of the payload in your solution? Because there is no initial velocity the equation below is applicable U1 * = () + (.1)2 o How did you analyze the horizontal motion of the payload in your solution? Because there is no horizontal force the equation below is applicable. o What other kinematics principles did you consider in analyzing the motion of the payload? I only used what was learned in the first for chapters MODIFIED SCENARIO ONE Scenario 1: In order to prepare pilots for changes in condition, I have been asked to model supply drops by a plane flying into a headwind. For this scenario, I will use my own location of Scio, Oregon. DiagramHead wind of WWW at 5 mats 2.2152 m/s in Scio Diegon V . 250 mph = 101.736my/ H -2650 m .Drop Zone Description A headwind is defined as a wind that's blowing directly in front of a forward motion in the opposite direction. Using the same information we have above for the velocity and height we have: The plane is flying at a constant speed of 250 mph and is 2650 meters above the ground. v = Velocity of the plane v = 250mph which = 111.736 m/s h = height of the plane h = 2650 m 1 mile = 1.609 km 1.609 km x 250 miles= 402 25 km mile hour hr Initial Horizontal velocity = 402.25 km/h = 111.736 m/'s 16 m = 111.736 m/s 402.25 The current wind speed in Scio, OR is WNW at 5pmh (US Department of Commerce, National Weather Service) 5mph = 2.2352 m/sThe vertical velocity stays the same with no change due to headwind t = time to hit the ground t = 23.2555 Using the equation R = v( x) o we have = 109.52 * 23.25 = 2546.34m = 2.54634 km 7 MODIFIED SCENARIO TWO Scenario 2: Due to historical difficulty in delivering supplies by plane, a colleague has suggested I develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees.MODIFIED SCENARIO TWO Scenario 2: Due to historical difficulty in delivering supplies by plane, a colleague has suggested I develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Catapult : Com below the forget site Launch arde is 5: Description The catapult is located at a fixed point of 400m away and 50m below the drop zone. it is capable of launching the payload at 67m/'s in an initial launch angle of 50V VDistance = 400m Height = 50m Angle = 500 Velocity = 67m/s t = time V(x) = vCOS50 (v COS 50) =67 + 0 642km/s = 43.0676m/s v(y) = VSIN50 (v SIN 500) = 67 x 0.766m/s = 51.322 m/s t = 400m/v(x) =400m + 43.0676 = 9.2837 t =9.28775 Determine if the catapult would be sufficient to deliver the payload to the drop zone H = v(y)t + 24/ = 51.322m/s x 9.2877s = 476.663 =476.663 - ( 0.5) ( 9.8) ( 9.2877) = = 53.982 53.982 is greater than 50, meaning that it Is indeed sufficient enough to deliver the payload to the drop zone. How did the changing variables affect the vertical and horizontal motion of the payload? Changing the variables did have an affect on the motion of the payload. By adding the headwind, in the 1st scenario, there would need to be an adjustment to the drop site made. In the original scenario, the drop site would be 2598.28m and in scenario 1 the drop site would be at 2546.34 which is a difference of 51.94m. In scenario 2, the catapult was in a fixed position and didn't have to have a force of gravity to complete the launch, only the initial forces, 67m/s and 50 degree launching angle, were needed.What adjustments to delivery would need to be made to account for these changing variables? The plane 1would need to adjust its speed accordingly.r in order to account for the headwind so the payload could still be dropped in the specified area. For the catapult, increasing either the distance from the drop zone or the height that it 1was located below the drop zone could have been increased or decreased depending on the angle