Question

Part B: Launching a Projectile at an Angle (1) Erase the trajectory from part A (yellow eraser button). (2) Ensure that the initial height (h = 10 m) and initial speed (1 = 15 m s 1) remain equal to the quantities given in part A. (3) Position the projectile launcher so that it launches at an angle of 20°. (4) Fire the launcher (using the red fire button) and record the x-displacement by positioning the center of the target to whether the projectile hit the ground. Height of launcher, h = _10 Initial speed, v = _15 m/s_ y-displacement, Ay = ___10_ Measured x-displacement, Ax= _28.7 Angle of launcher, e 20 The initial velocity that the ball leaves the launcher with is the same in both part A and part B. Using this, you will re-calculate the expected x-displacement in part B, Axc. SHOW YOUR WORK CLEARLY ON SEPARATE PAPER. I WILL NOT GRADE UNTIDY, UNCLEAR. UNFORMATTED WORK, & Q2a) First, showing formulae, calculate the initial velocity components of the projectile Vix Vy Q2b) Starting with the proper kinematic formula and showing all methodology, calculate the travel time, At, for part B. Q2c) Showing the symbolic formula, calculate the x-displacement, Axc for part B now that you have the travel time. Q2d) Calculate the percentage difference between the measured x-displacement, Axs, and the calculated x-displacement, Axc. Q3) For PART A, if the initial velocity is increased, will the ball reach the ground in a time greater than, equal to, or less than the time you calculated? Explain your answer clearly (no calculations). (4) If this experiment is taken to the moon (g = 1.6 m/s), explain, discussing time, if and why the distance travelled should be larger, smaller or the same than the distance measured in PART B of this lab. (Clearly explain WITHOUT using any mathematical equations.)

Answer 1

2a)

v_x = v_i cos $\theta$ = 15 * cos 20 = 14.1 m/s

v_y = v_i sin $\theta$ = 15 * sin 20 = 5.13 m/s

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2b)

$\Delta$y = v_yt + 1/2gt²

10 = 5.13 t + 4.9t²

so,

4.9t² - 5.13t - 10 = 0

so,

t = 5.13 +/- 14.9 / 9.8

neglect the negative value

so,

t = 2.044 seconds

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2c)

$\Delta$x = v_xt

$\Delta$x = 28.8 m

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2d)

% difference = | 28.8 - 28.7| \ ( 28.8 + 28.7 / 2) * 100

% difference = 0.347 %

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3a)

if we increase the velocity, time of flight will increase

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3b)

if gravity is less, the projectile will have more time in air and hence range will be more.