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Cannonball People!

This article delves into the physics of projectile motion using a cannonball as a case study. Given its horizontal velocity (Vx) of 30 m/s, an initial vertical velocity (Vyi) of 10 m/s, and a gravitational acceleration (a) of -9.81 m/s², we solve for various parameters including time of flight, horizontal displacement (Δx), and maximum height (ymax). The analysis reveals the total time of flight is approximately 2.04 seconds, the horizontal distance covered is 61.16 meters, and the peak height reached is 5.19 meters, with the vertex coordinates calculated.

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Cannonball People!

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  1. Cannonball People!

  2. Created Values • Vx= 30 m/s • Vyi= 10 m/s • a= -9.81 m/s²

  3. Solving for time • Vyf=-Vyi • Vyf=Vyi+at • (-10 m/s)= (10 m/s) + (-9.81m/s²) t • (-20m/s)/(-9.81m/s²)=t 2.0387s=t

  4. Solving for Change in x • Vx= Δx/t • 30 m/s = Δx/2.0387s • (30m/s)(2.0387s)=Δx Δx= 61.161m

  5. Solve for Change in y • ymax= Vyi(t/2) + ½ a (t/2)² • ymax=(10m/s)(2.0387s/2)+½(-9.81m/s²)(2.0387s/2)² • ymax=10.1935m+(-4.9999) ymax= 5.1936m

  6. Vertex of the Hoop • ( ½ Δx, ymax) • (½ 61.161m, 5.1936) Vertex= (30.5805,5.1936)

  7. Diagram! Top: 5.19m high 30.58m over 61.16m from canon

  8. Summary • Vx= 30 m/s • Vyi= 10 m/s • a= -9.81 m/s² • 2.0387s=t • Δx= 61.161m • ymax= 5.1936m • Vertex= (30.5805,5.1936)

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