/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 A helicopter is flying horizonta... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 90

A helicopter is flying horizontally at \(8.0 \mathrm{m} / \mathrm{s}\) and an altitude of \(18 \mathrm{m}\) when a package of emergency medical supplies is ejected horizontally backward with a speed of \(12 \mathrm{m} / \mathrm{s}\) relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?

Short Answer

Expert verified
Answer: The horizontal distance between the helicopter and the package when the package hits the ground is approximately 23.04 m.

Step by step solution

01

Determine the actual horizontal velocity of the package

Since the package is ejected horizontally backward relative to the helicopter, we need to find the actual horizontal velocity of the package. To do this, subtract the package's relative velocity from the helicopter's velocity: v_package = v_helicopter - v_relative v_package = 8.0 m/s - 12.0 m/s v_package = -4.0 m/s The package's horizontal velocity is -4.0 m/s, which means it's moving backwards relative to the helicopter.
02

Calculate the time it takes for the package to fall to the ground

Since air resistance is neglected, the only force acting on the package is gravity. We will use the following kinematic equation to find the time it takes for the package to fall to the ground: y = y_initial + v_y_initial * t + (1/2) * a_y * t^2 Where y is the vertical displacement, y_initial is the initial altitude, v_y_initial is the initial vertical velocity (0 m/s in this case), a_y is the vertical acceleration due to gravity (-9.81 m/s^2), and t is the time. As the package is dropped from an altitude of 18 m, the equation becomes: 0 = 18 + 0 * t + (1/2) * (-9.81) * t^2 Solving this quadratic equation for t, we get: t 鈮� 1.92 s
03

Calculate the horizontal position of the helicopter and the package at the time t

Now that we know the time it takes for the package to fall, we can find the horizontal position of the helicopter and the package at that time. For the helicopter: x_helicopter = x_helicopter_initial + v_helicopter * t x_helicopter = 0 + 8.0 m/s * 1.92 s x_helicopter 鈮� 15.36 m For the package: x_package =x_package_initial + v_package * t x_package = 0 + (-4.0 m/s) * 1.92 s x_package 鈮� -7.68 m
04

Calculate the horizontal distance between the helicopter and the package

Now that we have the horizontal position of the helicopter and the package at the time the package reaches the ground, we can find the horizontal distance between them: horizontal_distance = x_helicopter - x_package horizontal_distance = 15.36 m - (-7.68 m) horizontal_distance 鈮� 23.04 m The horizontal distance between the helicopter and the package when the package hits the ground is approximately 23.04 m.

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