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Projectile motion physics question 11 please​

Projectile motion physics question 11 please​-example-1
User Avendael
by
6.9k points

1 Answer

4 votes

Answer:

tair = 11.5 s

vi = 60.3 m/s

Step-by-step explanation:

In the x direction:

Given:

x₀ = 0 m

x = 600 m

v₀ = v cos 30° = v √3/2

a = 0 m/s²

x = x₀ + v₀ t + ½ at²

600 = 0 + (v √3/2) t + ½ (0) t²

600 = √3/2 v t

1200/√3 = v t

In the y direction:

Given:

y₀ = 300 m

y = 0 m

v₀ = v sin 30° = v/2

a = -9.8 m/s²

y = y₀ + v₀ t + ½ at²

0 = 300 + (v/2) t + ½ (-9.8) t²

0 = 300 + 1/2 v t − 4.9t²

Substitute:

0 = 300 + 1/2 (1200/√3) − 4.9t²

0 = 300 + 600/√3 − 4.9t²

4.9t² = 300 + 600/√3

t = 11.5 s

Solve for v:

v t = 1200/√3

v = 60.3 m/s

Graph:

desmos.com/calculator/1syounu0cg

User Michael Gattuso
by
6.7k points