/*! 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 13 One strategy in a snowball fight... [FREE SOLUTION] | 91Ó°ÊÓ

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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of \(25.0 \mathrm{m} / \mathrm{s} .\) The first one is thrown at an angle of \(70.0^{\circ}\) with respect to the horizontal. (a) At what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?

Short Answer

Expert verified
The second snowball should be thrown at an angle of approximately \( 30.1^\circ \) or \( 59.9^\circ \). Choose the smaller one, 30.1 degrees. The second snowball should be thrown approximately 1.409 seconds after the first.

Step by step solution

01

Compute the Range For the First Snowball

To calculate the range (distance) where the first snowball will land, use the maximum range formula for a projectile which is \( R = \frac{v^{2}sin(2\theta)}{g} \). Here, \( v \) represents the speed at which the snowball is thrown which is 25.0 m/s, \( \theta \) is the angle of projection, which is 70.0 degrees, and \( g \) is the acceleration due to gravity, which is 9.81 m/sec^2.
02

Compute the Time to Reach the Range for The First Snowball

The time \( t \) it takes for the first snowball to hit the ground can be computed using the formula \( t = \frac{2v \sin \theta}{g} \). Substituting the known values, calculate the time it takes for the first snowball to reach the ground.
03

Compute the Angle For the Second Snowball

Since we want the second snowball to reach the same point as the first, they will cover the same distance, however, it will be thrown at a lower angle. Use the range formula, \( R = \frac{v^{2}sin(2\theta)}{g} \), but this time, solve for \( \theta \), which will yield two possible values. Choose the smaller value since the problem specifies a 'low angle'.
04

Calculate when to Throw the Second Snowball

The second snowball needs to be thrown a certain amount of seconds after the first so that it arrives at the same time as the first snowball. Calculate this by subtracting the time taken for the second snowball to reach the target from the time for the first snowball. The equation to compute this time interval is \( \Delta t = t_{1}-t_{2} \) where \( t_{1} \) is time taken for first snowball and \( t_{2} \) is time taken for second snowball.

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Key Concepts

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

Trajectory Calculation in Projectile Motion
Understanding the trajectory calculation in projectile motion is crucial for accurately predicting where an object will land after being launched. The core of this calculation lies in physics equations that consider factors such as velocity, angle of launch, and the acceleration due to gravity.

Imagine throwing a snowball: the path it takes before hitting the ground is its trajectory. This path is determined by two components of motion - horizontal and vertical. Horizontally, the snowball moves at a constant speed because gravity doesn’t affect it in this direction. Vertically, it’s another story; gravity pulls the snowball down, creating a curved path.

To calculate trajectory, we use two key equations: one for range (horizontal distance) given by \[\begin{equation} R = \frac{v^{2}sin(2\theta)}{g} \end{equation}\] and one for time of flight \[\begin{equation} t = \frac{2v \sin \theta}{g} \end{equation}\]. In these formulas, \(v\) represents the launch speed, \(\theta\) is the angle of projection, and \(g\) is the gravitational acceleration, typically \(9.81 m/s^2\). By solving these, you can predict where and when the snowball will land, which is exactly what's needed to plan that perfect snowball ambush.
The Physics of Snowball Throwing
The act of throwing a snowball is governed by the essential principles of projectile motion. When a snowball is thrown, it immediately has two components of motion: the forward velocity that you impart and the constant acceleration downward due to gravity.

The initial velocity can be broken down into horizontal (\[\begin{equation} v_x = v \cos \theta \end{equation}\] ) and vertical (\[\begin{equation} v_y = v \sin \theta \end{equation}\] ) components. The horizontal velocity remains constant throughout the flight of the snowball if we ignore air resistance, while the vertical velocity changes due to gravity. This results in a parabolic trajectory, hallmark of projectile motion.

When calculating the optimal snowball-throwing strategy, it's essential to use the right angle and initial speed to ensure the snowball lands at the desired location. By leveraging the physics principles, one can craft a surprising snowball attack that has both, a high-angled arcing snowball and a fast, low-angled second throw, timed perfectly to hit the target simultaneously.
Projectile Angle and Range
The angle at which a projectile is launched directly affects its range—the distance it will travel horizontally before touching the ground. A projectile, like a snowball, thrown with the same speed but at different angles will follow different trajectories and cover different ranges.

At 45 degrees, a projectile will achieve the maximum range since the horizontal and vertical velocity components are equal. As shown in our earlier example for a snowball throw, angles greater or lower than 45 degrees yield lesser distances.

For a fixed speed, if one snowball is thrown upwards at a steep angle (like 70 degrees), it will reach a high peak but travel a shorter horizontal distance. A second snowball thrown at a smaller angle will have a lower peak but could travel the same horizontal distance if launched at the correct angle.

Choosing the Right Angle

To find the angle that gives the same range for different launch angles, we return to the range formula \[\begin{equation} R = \frac{v^{2}sin(2\theta)}{g} \end{equation}\] and solve for \(\theta\). This procedure lets us find the angle needed to ensure the second snowball hits the same spot as the first.

Knowing how to manipulate the angle of launch is not only fun for snowball fights but also has practical applications in sports, military strategy, and even space missions, where precise control over the trajectory is essential.

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Most popular questions from this chapter

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