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By throwing a ball at an angle of \(45^{\circ}\), a girl can throw the ball a maximum horizontal distance \(R\) on a level field. How far can she throw the same ball vertically upward? Asstme her muscles give the ball the same speed in each case. (Is this assumption valid?)

Short Answer

Expert verified
The maximum distance she can throw the ball vertically upward is equal to half the maximum horizontal distance she can throw.

Step by step solution

01

Understanding the given

Understand the given scenario. The ball is thrown at an angle of \(45^{\circ}\) which maximizes the horizontal range \(R\). The initial speed with which the girl can throw is the same in all situations.
02

Applying the Kinematic Equation for Horizontal throw

Ol pic apply the kinematic equation for the horizontal throw. We have the horizontal range \(R\) as \(R = V^2/g\) where \(V\) is the initial speed of throwing and \(g\) is acceleration due to gravity. This is obtained by applying the principle of conservation of energy.
03

Use the equation from step 2 for Vertical throw

Since the speed with which she throws is the same in both cases, we can use the equation we obtained earlier for the vertical throw as well. The maximum vertical height \(H\) to which she can throw the ball up can be given by the formula \(H = V^2/2g\). This is because while going upward the final velocity of the ball becomes zero at maximum height, hence the equation of motion \(V^2 = u^2 + 2gs\) simplifies to \(u^2 = 2gs\).
04

Compare \(R\) and \(H\)

Substituting the value of \(V^2/g\) from step 2 in the equation from step 3, we find that \(H = R/2\).

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Key Concepts

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

Kinematic Equations
Kinematic equations are mathematical formulas used to describe the motion of objects. These equations relate different aspects of motion, such as velocity, acceleration, time, and displacement. In the context of projectile motion, they are essential for analyzing the trajectories of objects that are launched through the air.

When a projectile is thrown, two primary kinematic equations come into play:
  • For vertical motion: The formula is given by \(V^2 = u^2 + 2as\), where \(V\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.
  • For horizontal motion: The range \(R\) can be calculated using \(R = V^2/g\), where \(V\) is the initial speed, and \(g\) is the acceleration due to gravity.
These equations allow us to determine important quantities like maximum range, height, and time of flight. Understanding these relationships is crucial for solving problems involving projectile motion.
Maximum Range
The maximum range in projectile motion occurs when an object is launched at an angle of 45 degrees. This angle offers the perfect balance between the horizontal and vertical components of velocity.

To achieve this maximum range, we use the formula \(R = V^2/g\), where \(V\) is the initial velocity and \(g\) is acceleration due to gravity. At this optimal angle, the horizontal distance covered by the projectile is maximized, assuming no air resistance.

Why 45 degrees? This angle ensures that the projectile equally balances its upward lift and forward thrust. Therefore, more time is spent in the air, allowing the projectile to cover more ground before landing. This concept of maximizing range is often applied in sports, ballistics, and engineering to achieve the furthest possible distance with the given initial speed.
Vertical and Horizontal Motion
In projectile motion, understanding the separation of vertical and horizontal motion is key for accurate analysis. Each direction of motion functions independently yet simultaneously.

For vertical motion, the projectile experiences acceleration due to gravity. When a projectile is launched vertically, its motion is influenced by gravity, causing the speed to decrease as it rises, and ultimately, come to a stop at its peak. The formula \(H = V^2/2g\) helps us find the maximum height the projectile can achieve, where \(H\) is the height, \(V\) is the initial velocity, and \(g\) is acceleration due to gravity.

Conversely, horizontal motion is constant because no other forces typically act along this direction, presuming air resistance is negligible. The horizontal displacement is governed by the initial speed and the time the projectile spends in the air. Hence, these distinct motions, when combined, describe the overall trajectory of the projectile, offering insights into how far, how high, and how long it flies.

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Most popular questions from this chapter

The record distance in the sport of throwing cowpats is \(81.1 \mathrm{~m}\). This record toss was set by Steve Urner of the United States in 1981 . Assuming the initial launch angle was \(45^{\circ}\) and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time the projectile was in flight. (c) Qualitatively, how would the answers change if the launch angle were greater than \(45^{\circ}\) ? Explain.

The cye of a hurricane passes over Grand Bahama Island in a direction \(60.0^{\circ}\) north of west with a speed of \(41.0 \mathrm{~km} / \mathrm{h}\). Three hours later the course of the hur. ricane suddenly shifts due north, and its speed slows to \(25.0 \mathrm{~km} / \mathrm{h}\). How far from Grand Bahama is the hurricane \(4.50 \mathrm{~h}\) after it passes over the island?

A jet airliner moving initially at \(3.00 \times 10^{2} \mathrm{mi} / \mathrm{h}\) due east enters a region where the wind is blowing \(1.00\) \(\times 10^{2} \mathrm{mi} / \mathrm{h}\) in a direction \(30.0^{\circ}\) north of east. (a) Find the components of the velocity of the jet airliner relative to the air, \(\vec{v}_{\mu}\). (b) Find the components of the velocity of the air relative to Earth, \(\overrightarrow{\mathbf{v}}_{M \mathcal{E}}\). (c) Write an equation analo- gous to Equation \(3.16\) for the velocities \(\vec{v}_{j}, \vec{v}_{A B}\), and \(\vec{v}_{g E}\) (d) What is the speed and direction of the aircraft relative to the ground?

A quarterback takes the ball from the line of scrimmage, runs backwards for \(10.0\) yards, then runs sideways parallel to the line of scrimmage for \(15.0\) yards. At this point, he throws a \(50.0\) -yard forward pass straight downfield, perpendicular to the line of scrimmage. What is the magnitude of the football's resultant displacement?

A hunter wishes to cross a river that is \(1.5 \mathrm{~km}\) wide and flows with a speed of \(5.0 \mathrm{~km} / \mathrm{h}\) parallel to its banks. The hunter uses a small powerboat that moves at a maximum speed of \(12 \mathrm{~km} / \mathrm{h}\) with respect to the water. What is the minimum time necessary for crossing?

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