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Chapter 2: Problem 45

A body is projected downward at an angle of \(30.0^{\circ}\) with the horizontal from the top of a building \(170 \mathrm{~m}\) high. Its initial speed is \(40.0 \mathrm{~m} / \mathrm{s}\). (a) How long will it take before striking the ground? (b) How far from the foot of the building will it strike? (c) At what angle with the horizontal will it strike?

Short Answer

Expert verified
(a) 7.0 s; (b) 242.2 m; (c) 68掳 below horizontal.

Step by step solution

01

Break Down Initial Velocity into Components

First, decompose the initial velocity into horizontal and vertical components using trigonometry. The initial speed is given as \( v_0 = 40.0 \, \text{m/s} \), and the angle is \( 30.0^{\circ} \). \[ v_{0x} = v_0 \cos(30.0^{\circ}) \] \[ v_{0y} = v_0 \sin(30.0^{\circ}) \] After calculation, \( v_{0x} = 40.0 \times \frac{\sqrt{3}}{2} \approx 34.6 \, \text{m/s} \) and \( v_{0y} = 40.0 \times \frac{1}{2} = 20.0 \, \text{m/s} \). These are the horizontal and vertical components of the initial velocity.
02

Calculate Time of Flight

Use the vertical motion equation to determine the time taken to reach the ground. The height of the building is \( 170 \, \text{m} \) and we'll solve for the time \( t \) using the equation for vertical motion: \[ h = v_{0y} t + \frac{1}{2} g t^2 \] where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = -170 \, \text{m} \). Rewriting the equation: \[ 0 = 170 + 20t - \frac{1}{2} \times 9.8 \times t^2 \] This becomes \( 0 = 4.9t^2 - 20t - 170 \). Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -20 \), and \( c = -170 \), we find \( t \approx 7.0 \, \text{s} \).
03

Determine Horizontal Range

Calculate how far from the base of the building the object strikes the ground using its horizontal motion. As there is no horizontal acceleration, the horizontal distance \( x \) is given by: \[ x = v_{0x} \times t \] Plug in \( v_{0x} = 34.6 \, \text{m/s} \) and \( t = 7.0 \, \text{s} \): \( x = 34.6 \times 7.0 = 242.2 \, \text{m} \).
04

Find the Impact Angle with the Horizontal

To find the angle of impact, first determine the components of the final velocity right before striking the ground. The vertical component is affected by acceleration due to gravity: \[ v_{y} = v_{0y} + gt = 20.0 + 9.8 \times 7.0 \approx 88.6 \, \text{m/s} \]. The horizontal component remains \( v_{x} = 34.6 \, \text{m/s} \) as there is no horizontal acceleration. Use the inverse tangent function to find the angle \( \theta \): \[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{88.6}{34.6}\right) \] Solving for \( \theta \), we get approximately \( 68^{\circ} \) below the horizontal.

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Key Concepts

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

Initial Velocity Components
When dealing with projectile motion, a crucial first step is breaking down the initial velocity into two components: horizontal and vertical. In this exercise, the object is thrown with an initial speed of 40 m/s at an angle of 30 degrees below the horizontal. By using trigonometry, you can separate the initial speed into these two perpendicular components:- **Horizontal Component (\(v_{0x}\))**: This is the part of the initial velocity along the horizontal axis and is calculated as \(v_{0} \cos(\theta)\). Here, \(\theta\) is 30 degrees, so \(v_{0x} = 40 \times \cos(30) = 34.6 \, \text{m/s}\).- **Vertical Component (\(v_{0y}\))**: This is the component along the vertical axis and is found similarly using \(v_{0} \sin(\theta)\). Therefore, \(v_{0y} = 40 \times \sin(30) = 20.0 \, \text{m/s}\).These components are fundamental as they allow us to analyze the motion in the horizontal and vertical directions independently.
Time of Flight
To find out how long an object is in the air, or its time of flight, you have to examine the vertical motion. From the exercise, the building height is 170 m, and you must solve for time \(t\) using the vertical motion equation:\[ h = v_{0y} t + \frac{1}{2} g t^2 \]Here, \(h\) is the vertical displacement, so we set it to \(-170 \, \text{m}\) indicating a downward motion. Given that \(v_{0y}\) is 20 m/s and \(g\), the gravitational acceleration is 9.8 m/s虏, the equation simplifies to:\[ 0 = 4.9t^2 - 20t - 170 \]Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.9\), \(b = -20\), and \(c = -170\), results in a time of approximately 7.0 seconds.
Horizontal Range
The horizontal range of a projectile is the distance it travels along the horizontal axis before hitting the ground. Since there is no air resistance, the horizontal motion is uniform. To compute this distance, use the equation:\[ x = v_{0x} \times t \]Substitute the known values, \(v_{0x} = 34.6 \, \text{m/s}\) and \(t = 7.0 \, \text{s}\), into the equation to find:\[ x = 34.6 \times 7.0 = 242.2 \, \text{m} \]This result indicates how far from the base of the building the object lands.
Impact Angle
To determine the angle at which the projectile impacts the ground, consider both the horizontal and vertical components of its velocity just before hitting the surface. At impact, gravity affects the vertical component, while the horizontal component remains unchanged:- **Final Vertical Velocity (\(v_{y}\))**: Starting with the initial \(v_{y0}\) and factoring in gravity's effect over time, \(v_{y} = v_{0y} + gt = 20.0 + 9.8 \times 7.0 \approx 88.6 \, \text{m/s}\).- **Horizontal Velocity (\(v_{x}\))**: Since there's no horizontal acceleration, \(v_{x}\) stays at its initial value, so \(v_{x} = 34.6 \, \text{m/s}\).With these velocities, you can calculate the impact angle \(\theta\) using:\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) = \tan^{-1}\left(\frac{88.6}{34.6}\right) \]After solving, the impact angle is approximately 68 degrees below the horizontal. This angle provides insight into the direction of movement as it strikes the ground.

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Most popular questions from this chapter

A marble dropped from a bridge strikes the water in \(5.0 \mathrm{~s}\). Calculate ( \(a\) ) the speed with which it strikes and \((b)\) the height of the bridge.

(a) Find the range \(x\) of a gun that fires a shell with muzzle velocity \(u\) at an angle of elevation \(\theta\). ( \(b\) ) Find the angle of elevation \(\theta\) of a gun that fires a shell with a muzzle velocity of \(120 \mathrm{~m} / \mathrm{s}\) and hits a target on the same level but \(1300 \mathrm{~m}\) distant. (See \(\underline{\text { Fig. } 2-8 .}\) ) (a) Let \(t\) be the time it takes the shell to hit the target. Then, \(x=v_{i x}\) tor \(t=x / v_{i x}\). Consider the vertical motion alone, and take \(u p\) as positive. When the shell strikes the target, $$ \text { Vertical displacement }=0=v_{i y} t+\frac{1}{2}(-g) t^{2} $$ Solving this equation gives \(t=2 v_{i y} / g .\) But \(t=x / v_{i x}\), so $$ \frac{x}{v_{i x}}=\frac{2 v_{i y}}{g} \quad \text { or } \quad x=\frac{2 v_{i x} v_{i y}}{g}=\frac{2\left(v_{i} \cos \theta\right)\left(v_{i} \sin \theta\right)}{g} $$ wherein \(g\) is positive. The formula \(2 \sin \theta \cos \theta=\sin 2 \theta\) can be used to simplify this. After substitution, $$ x=\frac{v_{i}^{2} \sin 2 \theta}{g} \text { . } $$ The maximum range corresponds to \(\theta=45^{\circ}\), since \(\sin 2 \theta\) has a maximum value of 1 when \(2 \theta=90^{\circ}\) or \(\theta=45^{\circ}\). (b) From the range equation found in \((a)\), $$ \sin 2 \theta=\frac{g x}{v_{i}^{2}}=\frac{\left(9.18 \mathrm{~m} / \mathrm{s}^{2}\right)(1300 \mathrm{~m})}{(120 \mathrm{~m} / \mathrm{s})^{2}}=0.886 $$ Therefore, \(2 \theta=\arcsin 0.886=62^{\circ}\) and so \(\theta=31^{\circ}\).

A skier starts from rest and slides down a mountainside along a straight descending path \(9.0 \mathrm{~m}\) long in \(3.0 \mathrm{~s}\). In what time after starting will the skier acquire a speed of \(24 \mathrm{~m} / \mathrm{s}\) ? Assume that the acceleration is constant and the entire run is straight, at a fixed incline, and around \(1.0\) -km long. We must find the skier's acceleration from the data concerning the 3.0 s trip. Taking the direction of motion down the inclined path as the \(+x\) -direction, we have \(t=3.0 \mathrm{~s}, v_{i x}=0\), and \(x=9.0 \mathrm{~m} .\) Then \(x=v_{i x} t+\frac{1}{2} a t^{2}\) gives $$ a=\frac{2 x}{t^{2}}=\frac{18 \mathrm{~m}}{(3.0 \mathrm{~s})^{2}}=2.0 \mathrm{~m} / \mathrm{s}^{2} $$ We can now use this value of \(a\) for the longer trip, from the starting point to the place where \(v_{f x}=24 \mathrm{~m} / \mathrm{s}\). For this trip, \(v_{i x}=0\), \(u_{f x}=24 \mathrm{~m} / \mathrm{s}, a=2.0 \mathrm{~m} / \mathrm{s}^{2} .\) Then, from \(v_{f}=v_{i}+a t\) $$ t=\frac{v_{f x}-v_{i x}}{a}=\frac{24 \mathrm{~m} / \mathrm{s}}{2.0 \mathrm{~m} / \mathrm{s}^{2}}=12 \mathrm{~s} $$

A body projected upward from the level ground at an angle of \(50^{\circ}\) with the horizontal has an initial speed of \(40 \mathrm{~m} / \mathrm{s}\). \((a)\) How long will it take to hit the ground? \((b)\) How far from the starting point will it strike? (c) At what angle with the horizontal will it strike?

A hose lying on the ground shoots a stream of water upward at an angle of \(40^{\circ}\) to the horizontal. The speed of the water is \(20 \mathrm{~m} / \mathrm{s}\) as it leaves the hose. How high up will it strike a wall that is a horizontal distance of \(8.0 \mathrm{~m}\) away?
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