Computer Problem # 1, Spring 2011 At t = 0, a projectile is located at the origi

Question

Computer Problem # 1, Spring 2011
At t = 0, a projectile is located at the origin and has a velocity of 20 m/s at 40 ° above the horizontal. The profile of the ground surface it strikes can be approximated by the equation y = 0.4x – 0.006x2, where x and y are in meters. Determine (a) the approximate coordinates of the point where it hits the ground, (b) its velocity and direction when it hits the ground, (c) its highest point in flight and (d) the greatest distance above the ground. Plot both the projectile’s flight and the ground profile. Discuss your results. Use engineering problem solving format. No hand marks, computer only.

Explanation / Answer

What a fun problem! What we need to do is find the x and y of the projectile and compare with the y for the ground at this point and see if they are the same (or if the y of the ground is higher). The projectile goes up and then down. The ground falls away. Some physics: the vertical velocity of the projectile changes. The horizontal velocity of the projectile never changes (until it hits the ground). It is always 20 m/s cos (40). Vertical velocity = 20m/s sin (40) - gt = 12.856 m/s - 9.8 t m/s Vertical position = 12.856 t (m) - 4.9 t^2 (m) Horizontal position = 15.321 t (m) Projectile Ground time x y y 1 15.32088886 7.955752194 4.719977732 1.2 18.38506663 8.370902632 5.325962603 1.4 21.44924441 8.394053071 5.819277249 1.6 24.51342218 8.02520351 6.19992167 1.8 27.57759995 7.264353949 6.467895866 1.9 29.10968884 6.736929168 6.55963163 1.92 29.41610662 6.619684212 6.574598676 1.926 29.50803195 6.583746325 6.578869083 1.927 29.52335284 6.577722377 6.579570958 Here I have the time in the left. The next column is time times the constant horizontal component of the starting velocity. The third column is the time from the first column in the formula for vertical projectile position. The far right column is the y value from the ground equation using the x value from the second column. As we can see, the projectile meets the ground after about 1.927 s near 29.52, 6.58 b) If we substitute 1.927s into the projectile velocity equations we get vert -6.029 m/s and horiz 15.32m/s for a combined 16.46 m/s at 21.5 degrees below the horizontal as the impact speed and direction. c) highest point is when vert speed is zero this is when x = v init^2/2g= 8.43m d) highest above ground is at about 0.96 seconds and is found by subrtacting the y values. 0.9 3.226443013 0.92 3.233895898 0.94 3.238555486 0.96 3.240421775 0.98 3.239494768 1 3.235774462 1.02 3.229260859 1.04 3.219953957 1.06 3.207853759 So now you have to program all this and plot it. I hope this is a good explanation.

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