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Problem 4: (Section 6.7) Two cannonballs are fired, both with initial velocity 100 ft/sec. The first is fired at an angle of 30 and elevation of 20 feet. The second is fired at an angle of 40 at ground level.

Find the parametric equations for the motion of the first cannon ball. [use 32ft/s^2]

Find the parametric equations for the motion of the second cannon ball.

Which cannonball goes higher? [i.e., what are the max heights of the two cannonballs?]

[algebraically, show how you arrived at your conclusion]

Which cannonball goes farther? [i.e., find the time at impact for each cannon ball, then determine how far each cannonball flew (horizontal distance)]

Algebraically, show how you arrived at your conclusion.

Bonus: At what angle should the first cannonball be fired at so that it reaches the same height as the second ball? [keep fixed at 100] [algebraically, show how you arrived at your conclusion]

Bonus: At what velocity should the second cannonball be fired at so that it lands at the exact same spot as the first ball? [keep the angle, fixed at 40] [algebraically, show how you arrived at your conclusion]

Bonus: At what angle should the second cannonball be fired at so that it lands at the exact same spot as the second ball? [keep fixed at 100] [algebraically, show how you arrived at your conclusion]

Answer 1

a) s_x = u_x*t = (100*cos30^o)*t = 86.6t

s_y = u_y*t + (1/2)a_yt² = (100*sin30^o)t + (1/2)(-32)(t²) = 50t - 16t²

s = √[(s_x)² + (s_y)²] = √[256t⁴ -1600t³ +9999.56t²]

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b) s_x = u_x*t = (100*cos40^o)*t = 76.6t

s_y = u_y*t + (1/2)a_yt² = (100*sin30^o)t + (1/2)(-32)(t²) = 64.3t - 16t²

s = √[(s_x)² + (s_y)²] = √[256t⁴ -2057.6t³ +10002.05t²]

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c) v_y = u_y + a_yt

At the highest point,

v_y = 0

=> t = -u_y/a_y

For the first cannonball,

t = -(100*sin30^o)/(-32) = 1.56 s

s_y = u_y*t + (1/2)a_yt² = 50t - 16t² = 39.06 ft

Since, the first cannonball is launched from a height of 20ft,

H_max = 20ft + 39.06ft = 59.06 ft

For the second cannonball,

t = -(100*sin40^o)/(-32) = 2.01 s

s_y = u_y*t + (1/2)a_yt² = 64.3t - 16t² = 64.6 ft

H_max = 64.6 ft

Therefore, the second cannonball goes higher

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d)

For the first cannonball,

When the cannonball reaches the farthest point (i.e. reaches ground)

s_y = -20

=> 50t - 16t² = -20

=> 16t² -50t -20 =0

=> t = 3.48 s

s_x = 86.6t = 301.4 ft

Therefore, range of the first cannonball is 301.4 ft

For the second cannonball,

When the cannonball reaches the farthest point (i.e. reaches ground)

s_y = 0

=> 64.3t - 16t² = 0

=> t*(16t -64.3) =0

=> t = 4.02 s

s_x = 76.6t = 307.9 ft

Therefore, range of the first cannonball is 307.9 ft

Therefore, the second cannonball goes farther