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Chapter 2: Problem 46

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?

Short Answer

Expert verified
The water strikes the wall at approximately 5.373 meters high.

Step by step solution

01

Break down the initial velocity into components

The water is shot at an angle, so its velocity has horizontal and vertical components. Use the angle \(40^\circ\) to find these. The horizontal component \(v_x\) is \(v \cos(40^\circ)\) and the vertical component \(v_y\) is \(v \sin(40^\circ)\).Given: \(v = 20 \text{ m/s}\)\[ v_x = 20 \cos(40^\circ) \approx 15.32 \text{ m/s}\]\[ v_y = 20 \sin(40^\circ) \approx 12.86 \text{ m/s}\]
02

Determine the time to reach the wall

Find the time \(t\) it takes for the water to reach the wall by using the horizontal motion equation: \(x = v_x \cdot t\).Given: \(x = 8.0 \text{ m}\)\[8.0 = 15.32 \cdot t\]Solve for \(t\):\[t = \frac{8.0}{15.32} \approx 0.522 \text{ s}\]
03

Calculate the vertical position at time t

Use the vertical motion equation to find how high the water reaches when \(t = 0.522\,\text{s}\):\[y = v_y \cdot t - \frac{1}{2}gt^2\]where \(g = 9.81 \text{ m/s}^2\) (acceleration due to gravity).Substitute the values:\[y = 12.86 \cdot 0.522 - \frac{1}{2} \times 9.81 \times (0.522)^2\]\[y = 6.710 - 1.337 \approx 5.373 \text{ m}\]
04

Conclude the height of impact

After calculating the height from Step 3, conclude that the stream of water will strike the wall at a height of approximately 5.373 meters.

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Key Concepts

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

Velocity Components
When studying projectile motion, one of the first steps is to break down the motion into its components. This is important because projectiles, like the stream of water from a hose, move simultaneously in both horizontal and vertical directions. Think of it as having two separate motions happening at the same time.

  • Horizontal Component: The part of the velocity parallel to the ground.
  • Vertical Component: The part of the velocity that acts perpendicular to the ground.
To determine these components, you use trigonometric functions: the angle of projection and the initial velocity. For instance, if a projectile is shot at an angle \( \theta \), the horizontal component (\( v_x \)) can be found using \( v_x = v \cos(\theta) \). Similarly, the vertical component (\( v_y \)) is \( v_y = v \sin(\theta) \). These components will help you calculate how far and how high the projectile will go.
Horizontal Motion
The horizontal motion of a projectile is often simpler than its vertical motion because it does not involve acceleration if we neglect air resistance. This means that the horizontal velocity component remains constant.

  • Distance (x): The total horizontal distance covered by the projectile.
  • Time (t): The duration it takes for the projectile to cover that horizontal distance.
Horizontal motion is governed by the equation \( x = v_x \cdot t \), where \( x \) is the horizontal distance and \( v_x \) is the horizontal component of the velocity. By rearranging the equation to solve for time (\( t \)), one can find how long it takes the water to reach a wall or target a certain distance away.
Vertical Motion
Vertical motion is slightly more complex because it's influenced by gravitational acceleration. This means that even though a projectile may start with an upward velocity, gravity will slow it down, bring it to a temporary halt at its highest point, and then increase its downward speed.

  • Initial Vertical Velocity (\( v_{y_0} \)): This is the vertical component calculated from the angle of projection.
  • Acceleration due to Gravity (g): On Earth, this constant is approximately \( 9.81 \text{ m/s}^2 \).
Vertical motion is described by the equation \( y = v_{y_0} \cdot t - \frac{1}{2} g t^2 \). This equation allows you to calculate the vertical position (\( y \)) of a projectile at a given time (\( t \)). Understanding these factors is crucial for predicting how high the water will reach before gravity pulls it back down.
Angle of Projection
The angle of projection significantly affects how high and how far a projectile will go. Simply put, it's the angle at which a projectile is launched from the ground.

  • Influence on Range: The angle determines the horizontal distance a projectile can cover. A 45-degree angle often gives the maximum range when speed is constant and air resistance is negligible.
  • Influence on Height: Higher angles generally result in greater maximum heights reached by the projectile.
In the provided exercise, the water was projected at a 40-degree angle. With this angle, we compute the components of the velocity, which in turn help predict the water's trajectory. Small changes in this angle can lead to significant differences in both the height and distance achieved, making it a crucial factor in projectile calculations.

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Most popular questions from this chapter

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 stone is thrown straight upward and it rises to a maximum height of \(20 \mathrm{~m}\). With what speed was it thrown? Take up as the positive \(y\) -direction. The stone's velocity is zero at the top of its path. Then \(v_{f y}=0, y=20 \mathrm{~m}, a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\). (The minus sign arises because the acceleration due to gravity is always downward and we have taken up to be positive.) Use \(v_{s}^{2}=v_{i}^{2}+2 a y\) to find $$ v_{i y}=\sqrt{-2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(20 \mathrm{~m})}=20 \mathrm{~m} / \mathrm{s} $$ Alternative Method You can check your result using the fact that the peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 s\), and so \(v_{i}^{2}=-2 g_{y}\), or \(v_{i}^{2}=-2(-9,81 \mathrm{~m} / 3)(20\) and \(u_{i}=19.8 \mathrm{~m} / \mathrm{s}\), or to two significant figures, \(u_{i}=20 \mathrm{~m} / \mathrm{s}\).

A ball is dropped from rest at a height of \(50 \mathrm{~m}\) above the ground. (a) What is its speed just before it hits the ground? (b) How long does it take to reach the ground? If we can ignore air friction, the ball is uniformly accelerated until it reaches the ground. Its acceleration is downward and is \(9.81\) \(\mathrm{m} / \mathrm{s}^{2}\). Taking down as positive, we have for the trip: $$ y=50.0 \mathrm{~m} \quad a=9.81 \mathrm{~m} / \mathrm{s}^{2} \quad v_{i}=0 $$ (a) \(v_{f y}^{2}=v_{i y}^{2}+2 a y=0+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(50.0 \mathrm{~m})=981 \mathrm{~m}^{2} / \mathrm{s}^{2}\) and so \(v_{f}=31.3 \mathrm{~m} / \mathrm{s}\). (b) From \(a\left(v_{f y}-v_{i y}\right) / t\), $$ t=\frac{v_{f y}-v_{i y}}{a}=\frac{(31.3-0) \mathrm{m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.19 \mathrm{~s} $$ (We could just as well have taken up as positive. How would the calculation have been changed?)

A baseball is thrown straight upward on the Moon with an initial speed of \(35 \mathrm{~m} / \mathrm{s}\). Compute \((a)\) the maximum height reached by the ball, (b) the time taken to reach that height, \((c)\) its velocity \(30 \mathrm{~s}\) after it is thrown, and \((d)\) when the ball's height is \(100 \mathrm{~m}\). Take up as positive. At the highest point, the ball's velocity is zero. (a) From \(v_{f y}^{2}=v_{i y}^{2}+2 a y\), since \(g=1.60 \mathrm{~m} / \mathrm{s}^{2}\) on the Moon, $$ 0=(35 \mathrm{~m} / \mathrm{s})^{2}+2\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=0.38 \mathrm{~km} $$ (b) From \(v_{f y}=v_{i y}+a t\) $$ 0=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t \quad \text { or } \quad t=22 \mathrm{~s} $$ (c) From \(v_{f y}=v_{i y}+\) at $$ v_{f j}=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(30 \mathrm{~s}) \quad \text { or } \quad v_{f y}=-13 \mathrm{~m} / \mathrm{s} $$ Because \(v_{f}\) is negative and we are taking up as positive, the velocity is directed downward. The ball is on its way down at \(t\) \(=30 \mathrm{~s} .\) (d) From \(y=v_{i y} t+\frac{1}{2} a t^{2}\) we have $$ 100 \mathrm{~m}=(34 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \quad \text { or } \quad 0.80 t^{2}-35 t+100=0 $$ By use of the quadratic formula, Or $$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ t &=\frac{35 \pm \sqrt{35^{2}-4(0.80) 100}}{2(0.80)}=\frac{35 \pm 30.08}{1.60} \end{aligned} $$ we find \(t=3.1 \mathrm{~s}\) and \(41 \mathrm{~s}\). At \(t=3.1 \mathrm{~s}\) the ball is at \(100 \mathrm{~m}\) and ascending; at \(t=41 \mathrm{~s}\) it is at the same height but descending.

A marble, rolling with speed \(20 \mathrm{~cm} / \mathrm{s}\), rolls off the edge of a table that is \(80 \mathrm{~cm}\) high. (a) How long does it take to drop to the floor? (b) How far, horizontally, from the table edge does the marble strike the floor?
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