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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards