Question

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?

Answer 1

6. 49.2 m/s

7. 9.18 s

8. 184 m

Step-by-step 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