Projectile Motion

Projectiles launched at an angle Start Stop

Projectiles launched at an angle vyf vy vx vx vy Δy θ vi Start Stop Δx

Projectiles launched at an angle Δx = vi(cos )Δt (horizontal distance) vyf = vi(sin ) – gΔt (final vertical velocity) vyf2= vi2(sin )2 – 2gΔy (final vertical velocity) Δy = vi(sin )Δt – ½ gΔt2 (vertical distance) if you are finding time with either of these two formulas, you are only covering half of the parabola

A projectile was launched at an angle of 22 A projectile was launched at an angle of 22.0o with an initial velocity of 25.0 m/s. a. How high will it go?

b. How long will it take to reach the top of the flight?

c. How long did it take the projectile to land?

d. How far down the field did the projectile travel?