THE EXPERIMENT For this experiment, we are going to use projectile motion simulation at h_ttps://www.gmgebra.org/m/aaymj In Part I, the initial speed \"v0\" of a ball will be measured by projecting the ball horizontally (9 : 0\") from at a height, yo, above the oor. In Part II, the ball will be projected at an initial angle 9. The horizontal displacement Dx, when the ball hits the ground (y = 0), and compared with the result from the simulation. A successful prediction with the predicted horizontal displacement for the angled projection, means that the following concepts about projectile motion are valid: 1. The horizontal component of the motion is independent of the vertical component in the two-dimensional motion; 2. Since there is no net horizontal force acting on the projectile, there is no horizontal component of the acceleration. This provides an experimental proof of Newton's 2"d law. 3. If ax : 0, once motion is imparted to the projectile, it will continue to move such that the horizontal component of its velocity is constant in time. This is also a consequence of Newton's rst law: \"An object at rest remains at rest; while an object in motion will continue to move with the same velocity unless acted upon by a net (unbalanced) force.\" I. HORIZONTAL SHOT (0 = 0) The initial projecting speed, v0, can be measured by projecting the ball horizontally: Using Eq. 4, if the horizontal displacement x, the ball travels from the launch point to the oor (y = 0), the time of flight, t, are obtained, then v0 can be calculated. (vOX = V0 cos 6 becomes vo because 6 = O). The time t is not measured but can be obtained from the vertical displacement. Since the ball is shot horizontally, the initial speed in the vertical direction is zero (V0y = sin (0) = 0). Thus, knowing the initial height (yo), time of ight, t, can be obtained from the vertical component of the motion. Procedure: 1. Open the projectile motion simulation at Mps://www.ggebra.org/m/aaym=j 2. Click the \"Reset\" button to place the ball at the initial position (0. yo) 3. Read and record yO from the simulation. 4. Set the initial angle to be zero, and click the \"run\" button to launch the ball. 5. Obtain the horizontal displacement, Dx, when the ball hits the \"ground\" (where y = 0) from the simulation. 6. From equation 6, obtain the time of ight, I. 7. From equation 4, obtain the initial speed, v0 8. Since the initial speed, V0, obtained will be used in the next section, it is best to repeat your readings, and calculations to be sure you have a consistent value for your initial speed.\\ Data Table 1: yo EX 2' V0 CONCLUSION State and discuss all concepts in projectile motion that were verified in this experiment. 2PROJECTILE MOTION PURPOSE 1.To investigate projectile motion in two dimensions. 2T0 verify that the horizontal component of the motion is independent of the vertical component. 3T0 measure the launch velocity of a projectile experimentally. 4.To apply the equations of motion to determine the path of the projectile at any instant of time. If the projectile hits a target, then theory is conrmed experimentally. BACKGROUND For a particle that is projected with initial velocity V0 at some angle 9 with the horizontal, the x and y components of the velocity are: V = v0 cos 9 Eq. 1(a) 0X v : V0 sin 9 Eq. 1(b) 0)' According to Newton's 2nd Law, Fnct = m a Eq. 2 which means (Fm)y : m ay Eq. 2(a) (Fm)X : m a)( Eq. 2(b) Neglecting air resistance, the only force acting on the particle in free ight is the force of gravity, which is directed downwards. This is a constant force and according to Eq. 2(a), the y component of the acceleration ay is constant. The value of cry is g (9.8 m/sl) downward. There is no net horizontal force after projecting the object; (F net)X = 0 and from Eq 2(b), aX = 0. In twodimensional motion of an object, the xcomponent of its motion is independent of the ycomponent. This concept and the equations of motion will be tested from the data obtained in this experiment. For projectile motion, the position as a function of time in two dimensions is given by r=r0+v0t+'/2at2 Eq.3 where r is the displacement of the particle relative to the origin (0,0) of the coordinate system. If the origin is chosen at the initial projecting point, r0 = 0. Since the horizontal acceleration ax = 0, the horizontal displacement of the particle becomes x=v0xt=vu cosGt Eq.4 and the horizontal velocity is Vx = v0 c056 = v0X Eq. 5 For the vertical displacement of the particle, y:voyt-'/2gt2:vosin9t-1/2th Eq.6 The vertical velocity is given by Vy :VOygt: Vosine-gt Eq.7 Equations 4, 5, 6, and 7 describe the motion of the projectile in tWO dimenSions' _ II. ANGLE SHOT The simulation is designed such that the initial speed of the ball is the same irrespective of the projection angle. In this section we get to review the concepts of projectile motion learned, this time, with an initial projection angle 8. Procedure: 1. From the simulation, choose an initial angle, I], using the initial angle slider in the simulation. 2. Using the initial speed v of the ball obtained in Part I, the '1 07 initial height ya, the chosen initial angle [I and the equations for the vertical component of the motion, calculate the time of ight. 3. Calculate, t_o predict the horizontal displacement, EX, where the ball will hit the \"ground\" (where y = 0). 4. Now, click the \"Reset\" button to place the ball at the initial position (0, yo) 5. Read and record y) from the simulation. 6. Click the \"run\" button to launch the ball. 7. Obtain the horizontal displacement, Dx, where the ball hits the \"ground\" (where y = 0) from the simulation. 8. Compare your predicted horizontal displacement, x, with that obtained from step 7 above, by doing a percent difference calculation. 9. Repeat steps 1 through 8 for tw_o more angles. Data Table 2: Quantity First trial:|:||:||:||:||:||] Second trial:|:|[|[||:||] Yo t Dxpmdima Dxcxpmmrnml QUESTION A ball A is thrown straight upwards and reaches its highest point in 1.5 s. A second ball B projected at an angle of 30 degrees above the horizontal, reaches the same maximum height as the rst ball. (a) At what speed was the first ball thrown? (b) At what speed was the second ball thrown? CONCLUSION In the report, state and discuss all concepts in projectile motion that were veried in this experiment. 2 3. Calculate, to predict the horizontal displacement, Ix, where the ball will hit the "ground" (where y = 0). 4. Now, click the "Reset" button to place the ball at the initial position (0, y.) 5. Read and record y. from the simulation. 6. Click the "run" button to launch the ball. 7. Obtain the horizontal displacement, [x, where the ball hits the "ground" (where y = 0) from the simulation. 8. Compare your predicted horizontal displacement, Ix, with that obtained from step 7 above, by doing a percent difference calculation. 9. Repeat steps 1 through 8 for two more angles. Data Table 2: Quantity First trial:00000[ Second trial:00000 Third trial:00000 yo t [X predicted X experimental %diff QUESTION A ball A is thrown straight upwards and reaches its highest point in 1.5 s. A second ball B projected at an angle of 30 degrees above the horizontal, reaches the same maximum height as the first ball. (a) At what speed was the first ball thrown? (b) At what speed was the second ball thrown?Lab 5 - PROJECTILE MOTION Data Sheet Name: ADate: I. HORIZONTAL SHOT (0 = 0") 1. Open the projectile motion simulation at https://Www.geogebra.org/m/aay