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Chapter 3: Problem 46

Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

Short Answer

Expert verified
The ratio of the height of the taller building to the shorter building is 4:1.

Step by step solution

01

Understanding the Problem

Two stones are thrown horizontally from two different buildings. The horizontal range of one stone is twice that of the other, implying a relationship between the heights of the buildings. The task is to determine the ratio of their heights.
02

Horizontal Motion Analysis

Let the velocity at which both stones are thrown be \( v_0 \). The horizontal distance covered by each stone (range) is given by \( R = v_0 \cdot t \) where \( t \) is the time of flight. Since one stone lands twice as far, if the range of the shorter building is \( R \), then the range for the taller building is \( 2R \).
03

Time of Flight Calculation

For horizontal motion, the time of flight \( t \) is determined by the time it takes for the stone to fall vertically. Using the equation for vertical motion \( h = \frac{1}{2}gt^2 \), solve for \( t \): \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height of the building and \( g \) is the acceleration due to gravity.
04

Equating for Range

Using the expression for time \( t \), the horizontal range \( R = v_0 \cdot \sqrt{\frac{2h}{g}} \). For the shorter building with height \( h_1 \) and range \( R \), the equation becomes \( R = v_0 \cdot \sqrt{\frac{2h_1}{g}} \). For the taller building with height \( h_2 \) and range \( 2R \), the equation is \( 2R = v_0 \cdot \sqrt{\frac{2h_2}{g}} \).
05

Setting Up the Ratio

Equate the expressions for \( R \) and \( 2R \): \( v_0 \cdot \sqrt{\frac{2h_1}{g}} = \frac{1}{2}v_0 \cdot \sqrt{\frac{2h_2}{g}} \). Simplify to find the relationship between \( h_1 \) and \( h_2 \).
06

Solving for the Height Ratio

Divide the two expressions to eliminate common terms and derive: \( \sqrt{h_2} = 2\sqrt{h_1} \). Squaring both sides gives \( h_2 = 4h_1 \). Therefore, the ratio \( \frac{h_2}{h_1} = 4 \).

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Key Concepts

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

Horizontal Range
The horizontal range in projectile motion refers to how far an object can travel horizontally before it hits the ground. In this particular exercise, both stones are thrown horizontally, meaning they have no initial vertical velocity.
  • This allows us to focus solely on horizontal motion when considering range.
  • The range is determined by the initial velocity of the stone and the time it is in the air.
The formula used for calculating horizontal range is given by:\[ R = v_0 \cdot t \]where \(R\) is the horizontal range, \(v_0\) is the horizontal component of initial velocity, and \(t\) is the time of flight.
In the original exercise, one stone lands twice as far from the base of its building as the other, leading to a ratio difference which is crucial in determining the heights of the two buildings.
Vertical Motion
Vertical motion in projectile motion occurs due to gravity acting on the object after it is thrown. In the case of stones thrown horizontally, the vertical motion begins right after they are released.
  • The vertical motion is separate from horizontal motion.
  • It solely depends on the height from which the stone is thrown and the acceleration due to gravity.
The equation governing vertical motion is:\[ h = \frac{1}{2}gt^2 \]where \(h\) is the vertical height, \(g\) is the acceleration due to gravity, and \(t\) is the time it takes to hit the ground.
The time of flight, \(t\), used in both horizontal and vertical calculations, connects the two motions and is crucial in comparing projectile distances.
Time of Flight
The time of flight is the duration the projectile spends in the air from the moment it is launched to the time it hits the ground. For horizontally launched projectiles, the time of flight is the same in both horizontal and vertical motions.
  • This time entirely depends on how long it takes for gravity to pull the stone down to the ground from its starting height.
The formula to find the time of flight from vertical motion is:\[ t = \sqrt{\frac{2h}{g}} \]where \(h\) is the height of the starting point, and \(g\) is the acceleration due to gravity.
This time of flight is a key factor in determining how far the projectile travels horizontally, as shown in the previous section on horizontal range.
Acceleration due to Gravity
In projectile motion, acceleration due to gravity (\(g\)) plays a critical role in influencing the path of the projectile. Gravity causes the projectile to accelerate downward, impacting both vertical motion and the time of flight.
  • The standard value for \(g\) is \(9.8 \, \text{m/s}^2\), but it can vary slightly depending on the location.
  • It affects the vertical motion predictably, ensuring all objects accelerate downwards at the same rate unless other forces act.
In horizontal projectile motion, gravity pulls the stone down, commencing its descent the moment it is released.
Gravity simplifies the calculation of the time of flight and affects how far horizontally the stone can travel, highlighting interdependencies within all aspects of projectile motion.

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Most popular questions from this chapter

Mario, a hockey player, is skating due south at a speed of \(7.0 \mathrm{m} / \mathrm{s}\) relative to the ice. A teammate passes the puck to him. The puck has a speed of \(11.0 \mathrm{m} / \mathrm{s}\) and is moving in a direction of \(22^{\circ}\) west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck"s velocity, as observed by Mario?

A projectile is launched from ground level at an angle of \(12.0^{\circ}\) above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced \(51.0 \mathrm{m}\) horizontally from the end of the ramp. His velocity, just before landing, is \(23.0 \mathrm{m} / \mathrm{s}\) and points in a direction \(43.0^{\circ}\) below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be \(162 \mathrm{km}\) away. The radar antenna is pointing upward at an angle of \(62.3^{\circ}\) from the ground. Find the \(x\) and \(y\) components (in \(\mathrm{km}\) ) of the position vector of the satellite, relative to the antenna.

A police officer is driving due north at a constant speed of \(29 \mathrm{m} / \mathrm{s}\) relative to the ground when she notices a truck on an east-west highway ahead of her, driving west at high speed. She finds that the truck's speed relative to her car is \(48 \mathrm{m} / \mathrm{s}\) (about \(110 \mathrm{mph}\) ). (a) Sketch the vector triangle that shows how the truck's velocity relative to the ground is related to the police car's velocity relative to the ground and to the truck's velocity relative to the police car. The sketch need not be to scale, but the velocity vectors should be oriented correctly and bear the appropriate labels. (b) What is the truck's speed, relative to the ground?
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