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Chapter 3: Problem 94

A plane diving with constant speed at an angle of \(49.0^{\circ}\) with the vertical, releases a package at an altitude of \(600 . \mathrm{m}\). The package hits the ground \(3.50 \mathrm{~s}\) after release. How far horizontally does the package travel?

Short Answer

Expert verified
Answer: The package travels approximately 359 meters horizontally before hitting the ground.

Step by step solution

01

Determine the angle and altitude in radians and meters respectively

First, convert the angle from degrees to radians since it is more suitable for calculations involving angles. Angle in radians = \((49.0^{\circ}) \times (\pi/180) = 0.855 \mathrm{rad}\) Now let's consider the altitude of the package. Altitude = \(600.0 \mathrm{m}\)
02

Determine the vertical velocity component

From the given angle, we can determine the vertical velocity component: \(V_v = V \times \sin(0.855)\) We need to find the value of V to calculate the vertical velocity component. To do this, we can use the equation for the time it takes for the package to hit the ground: \(t=3.50 \mathrm{s}\) For vertical motion, we know that \(s = ut + \frac{1}{2}at^2\) Using the vertical motion equation: \(600 = V_v \times 3.50 - \frac{1}{2} \times 9.81 \times (3.50)^2\) We can now solve for \(V_v\). \(V_v = \frac{600 + \frac{1}{2} \times 9.81 \times (3.50)^2}{3.50} = 183.495 \mathrm{m/s}\)
03

Determine the total velocity V

Now we can use the value of \(V_v\) to find the total velocity \(V\) as follows: \(V_v = V \times \sin(0.855)\) \(V = \frac{V_v}{\sin(0.855)} = \frac{183.495}{\sin(0.855)} = 202.152 \mathrm{m/s}\)
04

Determine the horizontal velocity component

With the total velocity \(V\), we can now calculate the horizontal velocity component: \(V_h = V \times \cos(0.855) = 202.152 \times \cos(0.855) = 102.563 \mathrm{m/s}\)
05

Calculate the horizontal distance

Finally, we can use the horizontal velocity component and the time it takes for the package to hit the ground to find the horizontal distance traveled: Horizontal Distance = \(V_h \times t = 102.563 \times 3.50 = 358.971 \mathrm{m}\) So the package travels approximately \(359 \mathrm{m}\) horizontally before hitting the ground.

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Most popular questions from this chapter

In ideal projectile motion, the velocity and acceleration of the projectile at its maximum height are, respectively, a) horizontal, vertical c) zero, zero. downward. d) zero, vertical downward. b) horizontal, zero. e) zero, horizontal.

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