which focuses on Moving Objects...
In a 2-dimensional motion (Projectile Motion),
there 2 basic components:
1.) Horizontal Motion
ax = 0 ; // horizontal acceleration is set to zero;
vx = v0x ; // horizontal speed is set as initial horizontal speed;
x = v0x . t ; // horizontal distance is the product of initial horizontal speed and time in seconds;
2.) Vertical Motion
ay = -g ; // vertical acceleration is equal to negative value of the gravitational force;
vy = v0y - (g . t) ; // vertical speed;
y = v0y . t - (1/2 g . t2) ; // vertical distance;
----------
Basic Equations:
g = 9.8 m/s2 ; // Gravitational Force;
vy = (√2g . h ) ; // Vertical Speed;
v0 = R (√ g / 2h ) ; // Initial Speed;
R = v0 (√2h / g ) ; // Range or Horizontal Distance;
h = ( 1/2 g . t2 ) ; // Height or Vertical Distance;
t = (√2h / g ) ; // Time or Duration;
R = Δ x = x - x0 = v0x . t = (v0x . t ) + (1/2 ax . t2 ) ; // Range;
h = Δ y = y - y0 = v0y . t - (1/2 g . t2) = (v0y . t ) + (1/2 ay . t2 ) ; // Height;
----------
Velocity-Initial:
V0x = ( V0 . cos θ );
V0y = ( V0 . sin θ ) - ( g . t );
Velocity-Final:
Vx = ( V0 . cos θ0 ) = V0x + ax . t ;
Vy = ( V0 . sin θ0 ) = V0y + ay . t ;
----------
Velocity of Projectile:
And after solving the x and y components, you must get the actual value for velocity (speed)
by using Pythagorean Theorem: v2 = vx2 + vy2;
V = √ Vx^2 + Vy^2 ; // Formula in solving for : velocity (speed) of a projectile;
----------
Angle of Projectile:
tan θ = (vy / vx ) ;
θ = tan-1 (vy / vx ) ; // Formula in solving for : Angle of Inclination;
Note : Cotangent is : cot or tan-1 ...
----------
Max or Peak:
ypeak = ( V0 . sin θ0 ) . tpeak - ( 1/2 g . ( tpeak )2 ) ;
Δ x = ( 2 V02 . [sin θ0 ] . [cos θ0 ] ) / ( g ) ;
Δ ypeak = ( 2 V02 . [sin θ0 ] 2 ) / ( 2 g ) ;
tpeak = ( V0 . sin θ0 ) / ( g ) ;
----------
Time:
t = 2(tpeak) = ( 2 V0 . sin θ0 ) / (g) ;
----------
----------
No comments:
Post a Comment