Respuesta :
Answer:
V = 308.1 m/s θ = -64º
Explanation:
a and b) We will use the projectile launch equations where we are asked to find the speed when the tank reaches the ground
Let's start by breaking down the speed
Vox = Vo cos θ
Voy = Vo sint θ
Vox = 140 cos 15 = 135.2 m / s
Voy = 140 sin 15 = 36.2 m / s
Let's look for vertical speed when it hits the ground
Vy² = Voy² - 2gy
Vy = √(36.2² - 2 9.8 3.98 103) = √ (1310-78008)
Vy = 276.9 m / s
We have both components.
V² = Vx² + Vy²
V = √ (135.2² + 276.9²)
V = 308.1 m / s
tan θ = Vy / Vx
tan θ = -276.9 / 135.2 = 2,048
θ = -64º
The negative sign means that it is measured from the x-axis clockwise
c and d) We repeat the same calculation for tank B, the only difference is the angle T = -15º
Vox = 140 cos (-15)
Voy = 140 sin (-15)
Vox = 135.2 m / s
Voy = 140 sin (-15)
Voy = -36.2 m / s
We calculate the vertical speed
Vy² = Voy² - 2 g Y
Vy = √ ((-36.2)² - 2 9.8 3980)
Vy = 276.9 m / s
V = √(135.3²2 + 276.9²)
V = 308.1 m/s
tan θ = -276.9 / 135.2
θ = -64
You can see that the speeds and angles are the same in both cases, the difference between these two situations is in the horizontal distance that runs each story