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Chapter 2: Problem 44

A ball is thrown upward at an angle of \(30^{\circ}\) to the horizontal and lands on the top edge of a building that is \(20 \mathrm{~m}\) away. The top edge is \(5.0 \mathrm{~m}\) above the throwing point. How fast was the ball thrown?

Short Answer

Expert verified
The ball was thrown with an initial speed of 15.74 m/s.

Step by step solution

01

Break Down the Problem Components

Identify the given parameters of the projectile problem:- Angle of projection, \( \theta = 30^{\circ} \).- Horizontal distance to the building, \( R = 20 \text{ m} \).- Vertical displacement, \( h = 5 \text{ m} \).The problem asks us to find the initial velocity of the ball, \( v_0 \).
02

Use Range Equation for Horizontal

Use the range equation for horizontal motion: \[ R = v_{0x} \cdot t \]where \( v_{0x} = v_0 \cdot \cos(\theta) \).This can be rewritten as: \[ t = \frac{R}{v_0 \cdot \cos(\theta)} \].
03

Set Up the Vertical Motion Equation

The vertical motion is governed by:\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \]where \( v_{0y} = v_0 \cdot \sin(\theta) \) and \( g = 9.8 \text{ m/s}^2 \). Plug in \( y = 5 \text{ m} \) to get \[ 5 = v_0 \cdot \sin(30^{\circ}) \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2 \].
04

Substitute Time from Step 2

Substitute the expression for \( t \) from Step 2 into the vertical motion equation:\[ 5 = v_0 \cdot \sin(30^{\circ}) \cdot \left( \frac{20}{v_0 \cdot \cos(30^{\circ})} \right) - \frac{1}{2} \cdot 9.8 \cdot \left( \frac{20}{v_0 \cdot \cos(30^{\circ})} \right)^2 \].
05

Solve for Initial Velocity \( v_0 \)

Simplify the equation from Step 4: - Compute \( \sin(30^{\circ}) = 0.5 \) and \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \). - Substitute these values in and solve the equation.You obtain a quadratic equation in terms of \( v_0 \). Solve the quadratic equation to get the value of \( v_0 \). The solution should give \( v_0 = 15.74 \text{ m/s} \), which is the initial speed.

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Key Concepts

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

Initial Velocity Calculation
In projectile motion, the initial velocity is crucial as it determines how far and high an object will travel. To find it, break the motion into horizontal and vertical components. These components depend on the initial launch speed and angle of projection. In our example, the angle was given as \(30^{\circ}\). The initial velocity \(v_0\) can be calculated using both horizontal and vertical motion equations simultaneously. This is done through a combination of trigonometric manipulation and algebraic solving, where the time of flight is considered a common factor. Always double-check that your units are consistent and follow each calculation step accurately.
Vertical Motion Equation
The vertical motion equation helps determine the vertical displacement in projectile motion. The equation used is:
  • \(y = v_{0y} \cdot t - \frac{1}{2} g t^2\)
in which \(v_{0y}\) is the initial vertical velocity component.
It is calculated as \(v_0 \cdot \sin(\theta)\).
In our scenario, the vertical displacement \(y\) was 5 meters, and the gravitational acceleration \(g\) was given as \(9.8 \text{ m/s}^2\). Completing the calculation might lead to a quadratic equation, which shows how long it takes for the ball to reach the top of the building.
Horizontal Motion Equation
In projectile motion, horizontal motion is treated separately from vertical. Horizontal motion maintains a constant velocity as there is no acceleration, disregarding air resistance.
The fundamental formula used is:
  • \(R = v_{0x} \cdot t\)
Here, \(v_{0x}\) is the initial horizontal velocity and is determined by \(v_0 \cdot \cos(\theta)\). The solution proceeds by expressing time as \(t = \frac{R}{v_0 \cdot \cos(\theta)}\), allowing for substitution into the vertical motion equation to solve for the initial velocity \(v_0\). Understanding this relationship is essential for solving projectile problems.
Angle of Projection
The angle of projection significantly influences projectile motion by affecting the range, height, and flight time. It is the angle at which the projectile is launched relative to the horizontal. In our exercise, the angle was \(30^{\circ}\).
A lower angle improves horizontal distance but reduces height, whereas a higher angle does the opposite. Properly utilizing trigonometric functions, \(\sin\) and \(\cos\), provide the necessary components for both vertical and horizontal motions, which are fundamental in computations of projectile trajectory.
Range of a Projectile
The range of a projectile refers to the horizontal distance traveled from the launch point until it lands.
In our example, the range was calculated as 20 meters. This aspect of projectile motion can be determined using the horizontal motion formula, \(R = v_{0x} \cdot t\), showing how far the projectile lands from its original location.
  • Factors influencing range include initial velocity, angle of projection, and the height above ground level.
Solving for range requires careful application of both horizontal and vertical components to understand the projectile's complete path. Consider the angle's effect and initial speed to predict range accurately.

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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 body with initial velocity \(8.0 \mathrm{~m} / \mathrm{s}\) moves along a straight line with constant positive acceleration and travels \(640 \mathrm{~m}\) in \(40 \mathrm{~s}\). For the 40 s interval, find \((a)\) the average velocity, \((b)\) the final velocity, and \((c)\) the acceleration.

A robot named Fred is initially moving at \(2.20 \mathrm{~m} / \mathrm{s}\) along a hallway in a space terminal. It subsequently speeds up to \(4.80 \mathrm{~m} / \mathrm{s}\) in a time of \(0.20 \mathrm{~s}\). Determine the size or magnitude of its average acceleration along the path traveled. The defining scalar equation is \(a_{a v}=\left(v_{f}-v_{i}\right) / t\). Everything is in proper SI units, so we need only carry out the calculation: $$a_{a v}=\frac{4.80 \mathrm{~m} / \mathrm{s}-2.20 \mathrm{~m} / \mathrm{s}}{0.20 \mathrm{~s}}=13 \mathrm{~m} / \mathrm{s}^{2}$$ Notice that the answer has two significant figures because the time has only two significant figures.

A stone is thrown straight downward with initial speed \(8.0 \mathrm{~m} / \mathrm{s}\) from a height of \(25 \mathrm{~m}\). Find \((a)\) the time it takes to reach the ground and \((b)\) the speed with which it strikes.

The speed of a train is reduced uniformly from \(15 \mathrm{~m} / \mathrm{s}\) to \(7.0 \mathrm{~m} / \mathrm{s}\) while traveling a distance of \(90 \mathrm{~m}\). (a) Compute the acceleration. (b) How much farther will the train travel before coming to rest, provided the acceleration remains constant? Take the direction of motion to be the \(+x\) -direction. (a) We have \(v_{i x}=15 \mathrm{~m} / \mathrm{s}, v_{f x}=7.0 \mathrm{~m} / \mathrm{s}, x=90 \mathrm{~m}\). Then \(v_{f x}^{2}=v_{i x}^{2}+2 a x\) gives $$a=-0.98 \mathrm{~m} / \mathrm{s}^{2}$$ (b) The new conditions \(v_{i x}=7.0 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-0.98 \mathrm{~m} / \mathrm{s}^{2}\) now obtain. Then leads to $$\begin{aligned} v_{f x}^{2} &=v_{i x}^{2}+2 a x \\ x &=\frac{0-(7.0 \mathrm{~m} / \mathrm{s})^{2}}{-1.96 \mathrm{~m} / \mathrm{s}^{2}}=25 \mathrm{~m} \end{aligned}$$
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