Question

Chapter 03, Problem 23 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 198 m/s, and each tank is at the same height of 1.73 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 from plane B. In each part, give the direction as a positive angle with respect to the horizontal. Fuel tank Pane A 15 Plane (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Answer 1

let vo = 198 m/s
h = 1.73 km = 1730 m
theta = 15 degrees

A) vox = 198*cos(15) = 191 m/s
voy = 198*sin(15) = 51.2 m/s

let t is the time taken for the tank to reach the ground.

use, -h = voy*t - (1/2)*g*t^2

-1730 = 51.2*t - (1/2)*9.8*t^2

on solving the above equation we get, t = 24.7 s

when the tank hits the ground, vx = vox = 191 m/s
vy = voy - g*t

= 51.2 - 9.8*24.7

= -191 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(191^2 + 191^2)

= 270 m/s <<<<<<<<<--------------Answer

B) direction : theta = tan^-1(vy/vx)

= tan^-1(191/191)

= 45 degrees <<<<<<<<<--------------Answer

C)

vox = 198*cos(15) = 191 m/s
voy = 198*sin(15) = -51.2 m/s

let t is the time taken for the tank to reach the ground.

use, -h = voy*t - (1/2)*g*t^2

-1730 = -51.2*t - (1/2)*9.8*t^2

on solving the above equation we get, t = 14.3 s

when the tank hits the ground, vx = vox = 191 m/s
vy = voy - g*t

= -51.2 - 9.8*14.3

= -191 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(191^2 + 191^2)

= 270 m/s <<<<<<<<<--------------Answer

D) direction : theta = tan^-1(vy/vx)

= tan^-1(191/191)

= 45 degrees <<<<<<<<<--------------Answer