91Ó°ÊÓ

As preparation for this problem, review Conceptual Example \(10 .\) The drawing shows two planes each about to drop an empty fuel tank. At the moment of release each plane has the same speed of \(135 \mathrm{~m} / \mathrm{s}\), and each tank is at the Plane \(\mathrm{B}\) same height of \(2.00 \mathrm{~km}\) 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 \(15.0^{\circ}\) above the horizontal and the other is flying at an angle of \(15.0^{\circ}\) below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A and (b) plane B. In each part, give the directional angles with respect to the horizontal.

Step by step solution

01

Resolve Initial Velocities

First, determine the horizontal and vertical components of the initial velocity for both planes. For Plane A, moving at an angle of \(15.0^{\circ}\) above horizontal, the components are \(v_{0x} = 135 \cos(15^{\circ})\) and \(v_{0y} = 135 \sin(15^{\circ})\). For Plane B, moving at an angle of \(15.0^{\circ}\) below horizontal, the components are \(v_{0x} = 135 \cos(15^{\circ})\) and \(v_{0y} = -135 \sin(15^{\circ})\).

02

Calculate Time of Flight

Since both tanks start from the same height, calculate the time it takes for them to hit the ground using the vertical motion equation: \( y = v_{0y}t + \frac{1}{2}gt^2 \). Here, \( y = -2000 \text{ m} \) (downward), and \( g = 9.81 \text{ m/s}^2 \). Use the quadratic formula to find \( t \).

03

Determine Final Vertical Velocity

Calculate the final vertical velocity \( v_{fy} \) at the moment they hit the ground using \( v_{fy} = v_{0y} + gt \). Substitute \( v_{0y} \) from each plane and the time \( t \) calculated in Step 2.

04

Calculate Magnitude of Final Velocity

Find the magnitude of the final velocity for both tanks using the Pythagorean theorem: \( v_f = \sqrt{v_{fx}^2 + v_{fy}^2} \), where \( v_{fx} = v_{0x} \) since horizontal velocity remains unchanged.

05

Calculate Direction of Final Velocity

Determine the directional angle with respect to the horizontal using \( \theta = \tan^{-1}\left(\frac{v_{fy}}{v_{fx}}\right) \). Consider the sign of \( v_{fy} \) to determine the direction with respect to the horizontal - negative for downward, positive for upward.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial velocity components

Understanding the initial velocity components is crucial in solving projectile motion problems. Even though both planes are flying at the same speed of 135 m/s, their velocities have different directions. This means we need to break down the velocity into horizontal and vertical components. For Plane A, with an angle of 15° above the horizontal, we calculate:

• Horizontal component: \( v_{0xA} = 135 \cos(15^{\circ}) \)
• Vertical component: \( v_{0yA} = 135 \sin(15^{\circ}) \)

Plane B flies at 15° below the horizontal, affecting its vertical direction. Thus:

• Horizontal component: \( v_{0xB} = 135 \cos(15^{\circ}) \)
• Vertical component: \( v_{0yB} = -135 \sin(15^{\circ}) \) (negative due to the downward direction)

These components are vital for further calculations in projectile motion.

Time of flight calculation

Calculating the time of flight is an essential step in determining when the projectile, in this case, the fuel tank, reaches the ground. Since both tanks are dropped from 2.00 km height, we'll use the equation for vertical motion:
\[ y = v_{0y}t + \frac{1}{2}gt^2 \]Here, \( y = -2000 \) meters (since the motion is downward), and \( g = 9.81 \text{ m/s}^2 \). This equation is quadratic in form, and solving it with the quadratic formula gives us the time, \( t \). Remember, the symmetrical properties of projectile motion apply here, allowing us to see how gravity affects the downward trajectory regardless of initial vertical speed.

Final velocity magnitude

After determining the time of flight, we can calculate the final velocity of the fuel tanks when they hit the ground. Each tank's vertical velocity component \( v_{fy} \) at impact can be found using:\[ v_{fy} = v_{0y} + gt \]The horizontal velocity remains unchanged, so it is still:\

• Plane A: \( v_{fx} = v_{0xA} \)
• Plane B: \( v_{fx} = v_{0xB} \)

We apply the Pythagorean theorem to compute the total final velocity magnitude,
\[ v_f = \sqrt{v_{fx}^2 + v_{fy}^2} \]For each plane, it's crucial to combine these components to see the overall impact speed of the tanks as they reach the ground.

Trajectory angle calculation

The trajectory angle is the last piece of our puzzle, giving us insight into the direction at which the fuel tanks strike the ground. Once the final velocity components are determined, we use the tangent function to find the angle \( \theta \) with respect to the horizontal:
\[ \theta = \tan^{-1}\left(\frac{v_{fy}}{v_{fx}}\right) \]The sign of \( v_{fy} \) dictates whether the angle is above or below the horizontal. Negative indicates a downward angle, aligning with the direction of gravity.
Understanding this angle helps in visualizing how projectile paths vary not only in speed but also in their downward path as they complete their journey toward the ground. It provides a complete picture of the tank's motion just before impact.