Question.5588 - Fill in all ___’s, and submit this worksheet for grading in Assignments Horizontal equations: (x-x0) = v0x t and vx = voxVertical equations: y = y0 + voy t + ½ a t2 and vy = voy + a t Component equations: v0y = vo cosθ v0y = vo sinθ Range equation: R = (vo2/g) sin(2θ) Pythagoras equation for vo, vo = vox2+voy2 , and angle equation θ = tan-1(voy/vox)https://www.compadre.org/Physlets/mechanics/illustration2_6.cfm Open the simulation, and run Animation 2 This is a ball being tossed straight upwards, to rise from a height of 0, and then it falls back down to a height of 0.Thus, the initial Vertical Velocity is not Zero. We begin using equations having “y” in our calculations.These are y = y0 + voy t + ½ a t2 vy = voy + a t (straight up and down means equations involving x are not needed) These equations use “up” as the positive direction, therefore gravity (being down) has a value of a = g = −9.8 m/s2. `Click on Animation 2 at the bottom, then click the run triangle. You will see the ball go up, then go down, and plots of y, vy, and a.1.a. Click on reset. The ball is at its starting height, here y0 = ____ m. Click run. The ball rises then drops to its final height, y = ____ m. Partly hidden, but legible beneath the ball, we can read the overall time for the up-and-down motion. This is t = ____ s.Symmetry has the time to go up = time to go down. Thus, tup = ____ s, and tdown = ____ s.1.b. We can use the y-equations above, to calculate the initial speed voy, and maximum height “ymax”. First, y is symmetric at t = 4 s, is this the max/min height (choose one). What is vy at this time? vy = ____ m/s.Now from the starting point, use vy = voy + a tup to calculate initial speed voy = _____ m/s.We now use y = y0 + voy tup + ½ a tup2 to calculate the max height at tup = 4 s, y = ________m. Click with your mouse on the upper, light-blue ball, and in yellow, you can approximately verify your max height calculation. 1.c. “Down trajectory half”. This last half accomplishes one thing--calculate its final velocity down--using tdown = 4 s. Its speed is zero when it begins its downward trajectory. Its initial height is y0 = final height in the “up” part, its voy = 0 and its final y = 0. Here, y0 =_____ m, vy = _______ m/s, (fill in y0, use voy and tdown to select an equation, and calculate final v). What has gone up has come back down, symmetrically on distance moved, antisymmetrically on velocity; all as time goes on, on each half of the full trajectory.Physlet Physics by Christian and Belloni: Illustration 3.4 (compadre.org) Open the simulation, and run. Projectile Trajectory.2. The overall time (end of simulation) is t = 8 s. So the left or “going up” half is 4 s, and the right or “going down” half is 4 s. Now we will put the cursor at various points, and clicking to get numbers (in the yellow box that comes up when you click). a. Click on the lower graph’s end point, to get the projectiles range, R = ______ m (even multiple of 10, first number in the yellow box).b. The upper-left blue graph shows unchanging horizontal velocity, click on it to get the value of vx. vox = vx = ______ m/s (about 30 m/s, second number in the yellow box).3. a. Click on upper-right vertical velocity graph, to measure its values (second number). The initial voy = _______ m/s. Final vy = −________ m/s (about + and – 40 m/s).b. We now verify the range equation, R = (vo2/g) sin(2θ). Start with vox = vx = 30 m/s, voy = 40 m/s values. Pythagoras gives us the initial vo, and tan-1(voy/vox) launch angle θ. Calculate, using these vox and voy values, vo = _______ m/s, θ = ______o.3. c. Finally use these calculated vo and θ values, in R = (vo2/g) sin(2θ) = _______ m. Is this close to measured R in 2a? Yes/No.Projectile Motion‬ (colorado.edu)‬‬‬‬‬‬‬‬‬‬‬‬‬‬ Open it, click on Intro, and check the following boxes: 4 Vector boxes, and lower “Slow” box.This is a horizontal initial velocity, projectile run. Click run (lower, red cannon box). You will see the green velocity vectors, and yellow acceleration vector. Green velocity components are horizontal and vertical down, and the yellow acceleration is down (single vector).4. Does the light green horizontal vector change its length? Yes/No. Does the green vertical (down) vector change its length? Yes/No (down arrow, within yellow arrow)Does the yellow acceleration vector change its length? Yes/No.The cannon’s height is indicated to its left. This height is “y0”. This is where we start our calculations. 5. Calculations: “horizontal initial velocity” projectile calculations have a strategy, first get the time-of-flight. This comes from the initial height yo, with final y = 0 (ground level). Here yo= 10 m, voy= 0, a = −9.8 m/s2. Then y=y0+voyt+ ½ a t2 gets t = _______ s (3 s.f.)6. We have the time-of-flight t, multiply the initial speed (below cannon) that never changes, to get range (or final x). x = ______ m.7. Now let’s try to hit the target. Move it to the right, to x = 25.0 m. Raise the cannon to yo = 12 m. What speed do we need? Remember time-of-flight is critical, and this comes from height yo. Calculate t = _______ s, for yo = 12 m (4 sig figs) Now we have a time-of-flight, and a distance x = 25 m that we need to cover. Use x − x0 = vox t to calculate vox = _______ m/s. Verify by setting the speed (nearest m/s) below the cannon and firing it. Are your stars rising, and does it check? Yes/No

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