To solve this problem, we use the formula
λ = s sin θ
where s is the separation and θ is the angle interference
So,
λ = 20 x 10^-6 sin 2.5
λ = 8.72 x 10^-7 m
The required angle for the fourth order bright fringe is
θb = sin⁻¹ (4λ / s) = sin⁻¹ (4 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 10.04°
The required angle for the fourth order dark fringe is
θd = sin⁻¹ (4.5 λ / s) = sin⁻¹ (4.5 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 11.31°
Answer:
414.9 m
Explanation:
First, become familiar with the horizontal, and vertical vector components.
Vertical vector: Vy = V × sin (θ).
Horizontal vector: Vx = V × cos(θ).
Distance traveled = Velocity vector × time in the air.
Time in the air given Vy = 2 × Vy / g (in respect to the metric of the vector).
Range of the projectile = Vx² / g
Time in the air given Vx = (Vx + √(Vx)² + 2gh) / g.
Given a 28° angle with an initial velocity of 70m/s, we have enough information to calculate!
Vx = 70 m/s × cos(28°) ≈ 61.806 m/s
Vy = 70 m/s × sin(28°) ≈ 32.863 m/s
t = 2 × Vy / g
t = 2 × ≈32.863 / 9.8
t = ≈65.726 / 9.8
t ≈ 6.7 s
Distance traveled (horizontal) = Vx × t = 61.806 × 6.7 ≈ 414.9 m
