Question

A Lockheed Martin F-35 II Lighting jet takes off from an aircraft carrier with a runway length of 90 m and a takeoff speed 70 m/s at the end of the runway. Jets are catapulted into airspace from the deck of an aircraft carrier with two sources of propulsion: the jet propulsion and the catapult. At the point of leaving the deck, the F-35’s acceleration decreases to a constant acceleration of 5.0m/s² at 30° with respect to the horizontal.

(a) What is the initial acceleration of the F-35 on the deck of the aircraft carrier to make it airborne?
a) 7.0 m/s²
b) 5.0 m/s²
c) 10.0 m/s²
d) 8.6 m/s²

(b) Write the position and velocity of the F-35 in unit vector notation from the point it leaves the deck of the aircraft carrier.
a) Position: ( (90hat{i} + 0hat{j}) ) m, Velocity: ( (70cos(30°)hat{i} + 70sin(30°)hat{j}) ) m/s
b) Position: ( (0hat{i} + 90hat{j}) ) m, Velocity: ( (70sin(30°)hat{i} + 70cos(30°)hat{j}) ) m/s
c) Position: ( (90hat{i} + 70hat{j}) ) m, Velocity: ( (70cos(30°)hat{i} + 70sin(30°)hat{j}) ) m/s
d) Position: ( (70hat{i} + 90hat{j}) ) m, Velocity: ( (70sin(30°)hat{i} + 70cos(30°)hat{j}) ) m/s

(c) At what altitude is the fighter 5.0 s after it leaves the deck of the aircraft carrier?
a) 115 m
b) 95 m
c) 125 m
d) 105 m

(d) What is its velocity and speed at this time?
a) Velocity: ( (35sqrt{3}hat{i} + 35hat{j}) ) m/s, Speed: ( 70 ) m/s
b) Velocity: ( (35hat{i} + 35sqrt{3}hat{j}) ) m/s, Speed: ( 70 ) m/s
c) Velocity: ( (35cos(30°)hat{i} + 35sin(30°)hat{j}) ) m/s, Speed: ( 70 ) m/s
d) Velocity: ( (35sin(30°)hat{i} + 35cos(30°)hat{j}) ) m/s, Speed: ( 70 ) m/s

(e) How far has it traveled horizontally?
a) 245 m
b) 210 m
c) 280 m
d) 175 m

Answer 1

(a) The initial acceleration of the F-35 on the deck of the aircraft carrier to make it airborne is (c) 10.0 m/s².

(b) The correct position and velocity of the F-35 in unit vector notation from the point it leaves the deck of the aircraft carrier is (c) Position: ((90hat{i} + 70hat{j})) m, Velocity: ((70cos(30°)hat{i} + 70sin(30°)hat{j})) m/s.

(c) The altitude of the fighter 5.0 s after it leaves the deck of the aircraft carrier is (a) 115 m.

(d) Its velocity and speed at this time are (c) Velocity: ((35cos(30°)hat{i} + 35sin(30°)hat{j})) m/s, Speed: (70) m/s.

(e) The horizontal distance traveled by the F-35 is (b) 210 m.

Step-by-step explanation:

(a) The initial acceleration of the F-35 can be calculated using the equation of motion:
`\(v^2 = u^2 + 2as\)`, where v is the final velocity (70 m/s), u is the initial velocity (0 m/s, as it starts from rest), a is the acceleration, and s is the distance (90 m). Solving for a, we get
`\(a = (v^2 - u^2)/(2s) = ((70)^2)/(2(90)) = 10.0 \, \text{m/s}^2\)`.

(b) The position vector at the given time is obtained using the kinematic equation
`\(s = ut + (1)/(2)at^2\)`, where u is the initial velocity, a is the acceleration, and t is time. The velocity vector is obtained by taking the derivative of the position vector concerning time.

(c) The altitude at a given time can be calculated using the vertical motion equation
`\(s = ut + (1)/(2)at^2\)`, where s is the altitude, u is the initial vertical velocity, a is the vertical acceleration, and t is time.

(d) The velocity vector is determined by combining the horizontal and vertical components using trigonometric functions. The speed is the magnitude of the velocity vector.

(e) The horizontal distance traveled is given by
`\(s = ut + (1)/(2)at^2\)`, where u is the initial horizontal velocity, a is the horizontal acceleration, and t is time.