Question

An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction 55${}^{\circ }$ above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?

Short answer

The pilot should release the hay 649 meters in front of the cattle.

Step-by-step solution

Analyze the Problem

We need to calculate the horizontal distance from the point of release to where the bales land. Use projectile motion because the bales are affected by gravity after being dropped.

Break Down Initial Velocity

Find the horizontal and vertical components of the initial velocity using trigonometry. Given the initial speed of 75 m/s, the horizontal component is $v_x=75\cos ({55}^{\circ })$ and the vertical component is $v_y=75\sin ({55}^{\circ })$.

Calculate Time of Flight

Use the vertical motion to find the time it takes for the bales to hit the ground. The initial vertical position is 150 m. Use the equation: $$y=v_{y}t+\frac{1}{2}g{t}^2$$where $g=-9.81{\text{ m/s}}^2$, solve for $t$ when $y=0$.

Solve for Time of Flight

Rearrange and solve the quadratic equation for time:$$-4.905t^2+75\sin ({55}^{\circ })t-150=0$$.Use the quadratic formula: $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to find $t$.

Find the Horizontal Distance

With the time of flight $t$, calculate the horizontal distance the bales travel using the horizontal velocity. Use:$x=v_x\times t$, where $v_x=75\cos ({55}^{\circ })$.

Final Calculation and Answer

Perform all calculations: 1. Calculate $v_x\approx 43.0\text{ m/s}$ and $v_y\approx 61.4\text{ m/s}$.2. Solve the quadratic equation to find $t\approx 15.1\text{ seconds}$.3. Calculate $x=43.0\times 15.1\approx 649\text{ meters}$.

Key concepts

Initial Velocity Components

When dealing with projectile motion, it's essential to break down the initial velocity into horizontal and vertical components. This allows us to analyze the motion separately in each direction.
For a projectile launched at an angle, the overall initial velocity (v) can be divided into two parts using simple trigonometry: the horizontal velocity (v_x) and the vertical velocity (v_y). This involves:

• Using cosine for horizontal motion: $v_x=v\cos (\theta )$
• Using sine for vertical motion: $v_y=v\sin (\theta )$

For our problem, where the airplane releases the bales with an initial speed of 75 m/s at 55°,

• The horizontal component: $v_x=75\cos ({55}^{\circ })\approx 43.0$ m/s
• The vertical component: $v_y=75\sin ({55}^{\circ })\approx 61.4$ m/s

Understanding these components is critical because they tell us how fast and in what direction the bales will initially start moving in both dimensions.

Time of Flight

The time of flight is the total time the bales are in the air, from the moment they leave the plane to when they hit the ground.
To calculate this, we focus on the vertical motion, since gravity only affects the projectile vertically.An important formula used here is the equation for vertical displacement:$$y=v_{y}t+\frac{1}{2}g{t}^2$$Where:

• $y$ is the vertical displacement, initially 150 m in our case.
• $v_y$ is the initial vertical velocity.
• $g$ is the acceleration due to gravity, $g=-9.81{\text{ m/s}}^2$.

For the bales to hit the ground from a height of 150 m, we solve when $y=0$.This results in a quadratic equation:$$-4.905t^2+61.4t-150=0$$Using the quadratic formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we solve for $t$, finding that the time of flight is approximately $15.1$ seconds.

Horizontal Distance Calculation

Once we know how long the bales are in the air, calculating the horizontal distance traveled is straightforward. The horizontal motion is constant because there's no horizontal acceleration acting on the bales.
The formula to find the horizontal distance ($x$) is given by:$$x=v_x\times t$$Where:

• $v_x$ is the horizontal component of the initial velocity, which we previously found to be $43.0\text{ m/s}$
• $t$ is the time of flight, $15.1$ seconds.

By substituting these values into the equation:$$x=43.0\times 15.1\approx 649\text{ meters}$$Thus, the pilot should release the bales approximately 649 meters before reaching the cattle, ensuring they land right where needed. Understanding this concept helps solve similar problems in projectile motion, ensuring a clear grasp of how motion progresses in two dimensions.