A projectile is fired with a horizontal velocity of 20 mfs and a vertical velocity of 45 6. What is the magnitude of its original velocity vector?
7. How long will the projectile be in the air?
8. What is its range?

6. 49.2 m/s

7. 9.18 s

8. 184 m

Explanation:

6. Use Pythagorean theorem.

v₀² = v₀ₓ² + v₀ᵧ²

v₀² = (20 m/s)² + (45 m/s)²

v₀ = 49.2 m/s

7. Given:

Δy = 0 m

v₀ᵧ = 45 m/s

aᵧ = -9.8 m/s²

Find: t

Δy = v₀ᵧ t + ½ aᵧt²

(0 m) = (45 m/s) t + ½ (-9.8 m/s²) t²

0 = 45t − 4.9t²

0 = t (45 − 4.9t)

t = 9.18 s

8. Given:

v₀ₓ = 20 m/s

aₓ = 0 m/s²

t = 9.18 s

Find: Δx

Δx = v₀ₓ t + ½ aₓt²

Δx = (20 m/s) (9.18 s) + ½ (0 m/s²) (9.18 s)²

Δx = 184 m