/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A boy can throw a ball a maximum... [FREE SOLUTION] | 91Ó°ÊÓ

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A boy can throw a ball a maximum horizontal distance of R on a level field. How far can he throw the same ball vertically upward? Assume that his muscles give the ball the same speed in each case.

Short Answer

Expert verified
The maximum vertical distance to which the boy can throw the ball is R/2

Step by step solution

01

Calculate initial speed

Using the given maximum horizontal distance, we first calculate the initial speed. From \( R = v^2 / g \), we solve for v to find \( v = \sqrt{Rg} \)
02

Calculate maximum vertical distance

We now substitute the initial speed \(\sqrt{Rg}\) into the equation for the maximum height \( H = v^2 / 2g \) to get the maximum vertical height to which the ball can be thrown: \( H = (\sqrt{Rg})^2 / 2g = R/2 \)

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Key Concepts

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

Initial Velocity Calculations
Understanding the initial velocity in projectile motion is crucial. It can help predict the path and distance a projectile will cover. In our exercise, a boy throws a ball, and we're tasked with finding out how high he can throw it given how far he can throw it horizontally.
To start, we use the maximum horizontal distance to calculate the initial velocity - the speed the ball leaves the boy's hand. The horizontal distance (R), the force of gravity (g, which is approximately 9.81 m/s²), and the initial velocity (v) are related.
We use the formula:
  • For horizontal distance:
    • \( R = \frac{v^2}{g} \)
    • Solving for v gives: \( v = \sqrt{Rg} \)
      This formula shows that initial velocity is proportional to the square root of the horizontal distance and gravity. Understanding this concept helps in many real-life applications, from sports to engineering.
Knowing the initial velocity allows us to further explore other components of projectile motion, like the horizontal range and vertical motion dynamics.
Horizontal Range
The horizontal range of a projectile is the maximum distance it covers along the horizontal axis. This distance depends on the initial velocity and angle of projection. Though our problem deals with a practical situation where the maximum distance is given, there are some underlying concepts worth understanding.
Typically, to achieve maximum horizontal range, the angle of projection should be 45°. This optimal angle lets the projectile reach the furthest point from its starting location. However, if that condition is met, the formula for calculating the range simplifies to:
  • \( R = \frac{v^2}{g} \)
    This formula stems from considering both the horizontal and vertical components of motion.
  • One crucial realization here is the impact of gravity as it pulls the projectile back to the ground. Gravity reduces the horizontal distance by dictating the time the projectile stays airborne.
When solving problems like the boy's throw, this formula helps us better comprehend how the initial velocity directly influences the projectile's horizontal journey.
Vertical Motion Dynamics
Vertical motion in projectile dynamics focuses on the changes in the vertical position of a projectile over time. In our exercise, once we have determined the initial speed from the horizontal throw, we can understand how it affects vertical motion.
The formula for maximum vertical height is derived from equating initial vertical kinetic energy to gravitational potential energy at the peak of motion:
  • \( H = \frac{v^2}{2g} \)
  • Substituting the initial velocity \( v = \sqrt{Rg} \), the formula simplifies to: \( H = \frac{R}{2} \)
    This conclusion elegantly shows that the maximum height the ball can reach is half of the maximum horizontal distance.
This relationship encapsulates an intriguing aspect of projectile motion where horizontal performance gives us insights into vertical potential. It emphasizes the symmetry in projectile paths and offers a clear visualization of the parabolic trajectory seen in real-world scenarios.

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Most popular questions from this chapter

The small archerfish (length 20 to \(25 \mathrm{cm}\) ) lives in brackish waters of southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of \(1.2 \mathrm{m}\) to \(1.5 \mathrm{m}\) and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target \(2.00 \mathrm{m}\) away, at an angle of \(30.0^{\circ}\) above the horizontal. With what velocity must the water stream be launched if it is not to drop more than \(3.00 \mathrm{cm}\) vertically on its path to the target?

In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor \(1.40 \mathrm{m}\) from the base of the counter. If the height of the counter is \(0.860 \mathrm{m},\) (a) with what velocity did the mug leave the counter, and (b) what was the direction of the mug's velocity just before it hit the floor?

The vector position of a particle varies in time according to the expression \(\mathbf{r}=\left(3.00 \hat{\mathbf{i}}-6.00 t^{2} \hat{\mathbf{j}}\right) \mathrm{m} .\) (a) Find expres- sions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at \(t=1.00 \mathrm{s}\)

A golf ball is hit off a tee at the edge of a cliff. Its \(x\) and \(y\) coordinates as functions of time are given by the following expressions: $$\begin{array}{c}x=(18.0 \mathrm{m} / \mathrm{s}) t \\ \text { and } \quad y=(4.00 \mathrm{m} / \mathrm{s}) t-\left(4.90 \mathrm{m} / \mathrm{s}^{2}\right) t^{2} \end{array}$$ (a) Write a vector expression for the ball's position as a function of time, using the unit vectors \(\mathbf{i}\) and \(\mathbf{j} .\) By taking derivatives, obtain expressions for (b) the velocity vector \(\mathbf{v}\) as a function of time and (c) the acceleration vector a as a function of time. Next use unit- vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the golf ball, all at \(t=3.00 \mathrm{s}\)

Heather in her Corvette accelerates at the rate of \((3.00 \hat{\mathbf{i}}-2.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2},\) while Jill in her Jaguar accelerates at \((1.00 \hat{\mathbf{i}}+3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) They both start from rest at the origin of an \(x y\) coordinate system. After \(5.00 \mathrm{s},\) (a) what is Heather's speed with respect to Jill, (b) how far apart are they, and (c) what is Heather's acceleration relative to Jill?

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