The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 140 m/s, and each tank is at the same height of 3.98 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0° above the horizontal and the other is flying at an angle of 15.0° below the horizontal. Find the (a) magnitude and (b) direction of the velocity with which the fuel tank hits the ground if it is from plane A. Find the (c) magnitude and (d) direction of the velocity with which the fuel tank hits the ground if it is from plane B. In each part, give the direction as a positive angle with respect to the horizontal.

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