I used to be good at phyics, but

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SUMMARY

The discussion revolves around calculating the trajectory of a cannonball fired at a 45-degree angle with an initial velocity of 500 m/s from a height of 5 feet. Key equations discussed include the kinematic equations for projectile motion, specifically the horizontal displacement formula x = v_x t and the vertical motion equation y = y_0 + v_{y0} t + (1/2) a t^2. The participants successfully derived the time of flight as approximately 8.52 seconds and the total horizontal distance traveled as approximately 3367.87 meters, while also addressing the maximum height reached by the projectile.

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  • Basic trigonometry for resolving velocity components
  • Knowledge of gravitational acceleration (9.81 m/s²)
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  • #31


xxChrisxx said:
You got to think, when does y max occur.

When Vy=0, Y is at its highest.

So,...
 
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  • #32


so you know the

initial velocity
final velocity
acceleration

you can use
v^2=u^2+2as to find the max height (remember to +5ft)

v=u+at to find time
 

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