22-Sept-2016: Trajectories

Apparatus: 
    Two aluminum "v-channel", a steel ball, a board, a ring stand, a clamp, some paper, a carbon paper.

Purpose: 
     To use your understanding of projectile motion to predict the impact point of a ball on an inclined board.

Model: 
    V0*sinθ1 = Vx, V0*cosθ1 =Vy, Δx1 = Vx*t1, Vy*t1+(gt1^2)/2 = h, tanθ2 =Δy2/Δx2
    V0 is the initial velocity of the ball when the ball left the channel, Vx is the velocity in x-axis, Vy is the velocity in y-axis, t1 is the time thai the ball hit the ground, Δx is the distant between the end of the channel and the hit point on the ground in x-axis, h is the heigh of the end of the channel, and θ2 is the angle of the board.

Process: 
    First, we set up the apparatus. We use the rind stand and a v-channel to make a incline and connect it to a horizontial channel to make a "launch trajectory" of the ball. We try to launch the ball once in order to find out the possible landing point, and then we put a carbon paper on the gound around the predicted landing point and cover a piece of paper on it.

    We release the ball on the incline at a same heigh five times. By determining the distant between the end of the channel and the hit point on the ground in x-axis, we get  Δx1 = 0.605m. By detemining the heigh of the end of the channel, we get h = 0.932m.
    By using the equation and calculation:
                                                          Vy*t1+(gt1^2)/2 = h
                                                          Δx1 = Vx*t1
                                                                             V0*sinθ1 = Vx
                                                                             V0*cosθ1 =Vy
                                                                              h = 0.932m
                                                          Δx1 = 0.605m
     We get:                                                                      
                                                                   t1 = 0.4365s, V0 = 1.387m/s

    Then, we put a board at the end of the channel to make anothor incline, and we use our cellphone to determind the angle of the board, θ2 = 44°.  We place a carbon paper and a paper on the incline, and then we releave the ball at the same position five times. We determine the distant between the end of the channel and the landing point d = 0.535m.
    By using the equation and calculation:
                                                          tanθ2 =Δy2/Δx2
                                                                             V0*sinθ1 = Vx
                                                          V0*cosθ1 =Vy
                                                          Δx2 = Vx*t2
                                                          Vy*t2+(gt2^2)/2 = Δy2
                                                                             D = Δx2/sinθ2
                                                              t1 = 0.4365s, V0 = 1.387m/s ,  θ2 = 44°
    We get:
                                                          D = 0.527

    We compare the result D that got by calculation and the result d that got by determining. We get two very similar number. We conside that there are some error when we determine. We think our experiment is success.