Two Dimensional Projectile Motion Y m Horizontal Projectile

Two Dimensional Projectile Motion Y (m) Horizontal Projectile --- Part 1 X (m) Time t is the link between X & Y

Horizontal Projectile Vx: constant H = ½ gt 2 Free Fall Voy=0

X=Vt (ax = 0, ay = g)

Measurements • • • Horizontal Distance X: X = (X 1 + X 2 +X 3 + X 4 + X 5) / 5 Vertical Distance Y: Y : the height of the gun Both X and Y are in meters • ( Measure 2 times: for Horizontal and 15 degree angle Projectiles )

Theoretical Calculation – Horizontal projectile • X = V t (no acceleration in X direction) • X is the average of 5 measurements of horizontal distance. • V=X/t • V is the initial velocity of the bullet – in X direction – a constant (no acceleration in X. . ) • V only depends on the character of the GUN, • It doesn’t change with its position.

Time t links X and Y directional motion • • V=X/t in X direction Y = ½ g t 2 in Y direction It is the same t, both in X and Y direction. If we can get t from the Y equation, substitute t to the X equation, we can calculate V the initial velocity.

Example: X = 2. 8 m, Y = 1 m • • Y = ½ g t 2 1 = ½ x 9. 8 t 2 = 1 / 4. 9 t = 0. 45 s Sub t = 0. 45 s to V = X / t V = 2. 8 / 0. 45 = 6. 2 m/s V is the initial velocity of the bullet, independent of its position, same for 15 degree angle

Projectile with an angle

Decompose V to Vx and Vy Vy Vx Vx = V cos 15 = 6. 2 x 0. 966 = 6. 0 m/s Vy = V sin 15 = 6. 2 x 0. 26 = 1. 6 m/s

again, t links X and Y motion • • X = Vx t in X direction -Y = Vy t – ½ g t 2 in Y direction time t is the same for X and Y direction measured Y = 1. 03 m ( height of the gun) -1. 03 = 1. 6 t – 4. 9 t 2 4. 9 t 2 – 1. 6 t -1. 03 = 0 t = 0. 65 s

Compare X --- error% • Sub t = 0. 65 s to the X equation, • X cal = Vx t = 6 x 0. 65 = 3. 9 m • X exp = (X 1 + X 2 + X 3 + X 4 + X 5) / 5 • X error% = (X cal – X exp) / X cal x 100%