physics word problem

Question

As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each dropping an empty fuel tank. At the moment of release each plane has the same speed of 145 m/s, and each tank is at the same height of 2.01 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.

Answer 1

This is a 2D kinematics problem. We need to separate everything into the horizontal direction (x) and the vertical direction (y). I use the subscript ₀ for initial (when the package is dropped) quantities and _f for final (when the package lands) quantities.The numbers given in the problem have 3 significant figures. I try to keep everything in variable form until we plug in numbers at the very end. Here is list of equations we need

v_fy² = v_0y² + 2a_yΔy

v_f = √(v_fx² + v_fy²)

θ_f = tan^-1(v_fy / v_fx)

cos²(θ) + sin²(θ) = 1

The speed is 145 m/s. Let's call this variable v₀. The angle is 15° or -15° which we will call θ₀. We want to calculate v_0x and v_0y which are the x and y components of the initial velocity. One of them is v₀sin(θ₀) and the other is v₀cos(θ₀).

I find it easiest to remember what happens at θ=0. We should memorize that sin(0) = 0 and cos(0) = 1. If the angle were 0 in this problem, the velocity would be entirely in the horizontal direction. Therefore v_0x=v₀cos(θ₀) and v_0y=v₀sin(θ₀).

We should also note that v_oy is positive when the angle is above the horizontal as in Plane A and negative for Plane B. This is satisfied if we say that θ_0A = 15° and θ_0B = -15º because sin(-θ) = -sin(θ). We will not separate cases A and B until the end when we plug in numbers for variables.

There is no acceleration in the x direction. The acceleration in the y direction is due to the force of gravity. The value of this acceleration is -9.8 m/s² which we will call -g. Therefore, the accelerations in each direction, which are constant, are a_x=0 and a_y=-g.

When there is no acceleration, there is no change in velocity. This means that the x component of the velocity is constant from when the package is released to when it lands. Use v_oy, a_y, and height (Δy) to determine v_y which is the y component of the velocity when the package hits the ground. Note that Δy = y_f - y₀ will be negative.

v_fx = v_0x = v₀cos(θ₀)

v_fy² = v_0y² + 2a_yΔy = [v₀sin(θ₀)]² + 2(-g)Δy

The problem asks for the magnitude (v_f) and direction (θ_f) of the final velocity.

v_f = √(v_fx² + v_fy²) = √([v₀cos(θ₀)]² + [v₀sin(θ₀)]² + 2(-g)Δy) = √(v₀²+ 2(-g)Δy)

θ_f = tan^-1(v_fy / v_fx) = tan^-1(√([v₀sin(θ₀)]² + 2(-g)Δy) / v₀cos(θ₀))

Finally, we can plug in v₀ = 145 m/s, g = 9.8 m/s², Δy = -2.01 km and θ₀ = 15º or -15º. Make sure to convert km to m and set you calculator to "degrees" instead of radians when you plug everything in. Here's what I got

v_f = 246 m/s

θ_f = 86.7°

Sanity and understanding check

Make sure you understand the answers to the following questions both mathematically and physically.

1. Why is v_f > v₀?
2. Why is θ_f > θ₀?
3. What happens when θ₀ = 0 or θ₀ = 90º?
4. What happens if we're in space with no gravity (g = 0)?
5. Why did we get the same answer for Plane A and Plane B?