/*! 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 77 In an action-adventure film, the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going \(90.0 \mathrm{km} / \mathrm{h},\) to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.

Short Answer

Expert verified
Initial velocity relative to hero: approximately 20 m/s. Velocity relative to Earth: around 39 m/s.

Step by step solution

01

Convert Speeds from km/h to m/s

Convert the speeds of both cars from km/h to m/s:- Hero's car: \[ v_{hero} = 90.0 \text{ km/h} = \frac{90.0 \times 1000}{3600} \approx 25 \text{ m/s} \]- Enemy's car: \[ v_{enemy} = 110 \text{ km/h} = \frac{110 \times 1000}{3600} \approx 30.56 \text{ m/s} \]
02

Determine Relative Velocity of Cars

The enemy's car is moving faster than the hero's car: \[v_{relative} = v_{enemy} - v_{hero} = 30.56 \text{ m/s} - 25 \text{ m/s} = 5.56 \text{ m/s}\] This is the relative speed gained by the enemy.
03

Use Relative Velocity to Plan Grenade Throw

Given that the grenade is thrown at an angle of \(45^\circ\) and needs to travel 15.8 m horizontally:The horizontal and vertical components of the initial velocity, \(v_{0x}\) and \(v_{0y}\), are equal due to the angle.- Horizontal component: \[ v_{0x} = v_0 \cos(45^\circ) = v_0 \cdot \frac{1}{\sqrt{2}} \]- Vertical component: \[ v_{0y} = v_0 \sin(45^\circ) = v_0 \cdot \frac{1}{\sqrt{2}} \]
04

Calculate Time to Pass the Distance

Calculate how long it takes for the grenade to cover the relative horizontal distance:Using the formula for relative motion, \( s = ut \):\[ 15.8 = v_{0x} \cdot t + v_{relative} \cdot t\]Substitute for \(v_{0x}\) using 45-degree projection:\[ 15.8 = \left(v_0 \cdot \frac{1}{\sqrt{2}}\right) \cdot t + 5.56 \cdot t\]Simplify to solve for \(t\):\[ 15.8 = t \left(\frac{v_0}{\sqrt{2}} + 5.56\right)\]\[t = \frac{15.8}{\frac{v_0}{\sqrt{2}} + 5.56}\]
05

Set Up Equations for Vertical Motion

For the vertical motion:The motion formula is\[ v_{0y} \cdot t - \frac{1}{2}gt^2 = 0\]where \(g = 9.81 \text{ m/s}^2\).Solve for \(t\):\[v_0 \cdot \frac{1}{\sqrt{2}} \cdot t = \frac{g}{2} \cdot t^2\]
06

Solve for Initial Velocity \(v_0\)

Combine the equations from Steps 4 and 5 and solve for \(v_0\):Since the vertical and horizontal components use the same time \(t\), solve:\[v_0^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = g \cdot \left(\frac{15.8}{\frac{v_0}{\sqrt{2}} + 5.56}\right)\]Simplify to get the quadratic equation and solve for \(v_0\):\( \sqrt{2}v_0 + 5.56 = \sqrt{2g}\) gives solution for \(v_0\).
07

Determine Velocity Relative to Earth

Find the velocity relative to the Earth after solving \(v_0\):The total velocity:\[v_{earth} = v_hero + v_{0x} = 25 + v_0 \cdot \frac{1}{\sqrt{2}}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Relative Velocity
Relative velocity is all about understanding how one object's speed is perceived from the frame of reference of another moving object. In our scenario with cars chasing each other in an action scene, the enemy car is moving faster than the hero car. The relative velocity is calculated as the difference in their speeds.

Since both cars are moving in the same direction, we subtract the speed of the hero's car from the enemy's car speed. This gives us the relative speed of 5.56 m/s that the enemy is gaining on the hero.

Knowing this relative velocity helps when planning any projectile motion, such as throwing a grenade from one moving car to another. It allows us to calculate how quickly the distance between the two cars is changing, affecting how far and fast the projectile needs to be thrown.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. In this problem, we focus on kinematics to predict the path and landing of the grenade.

The movement can be described in terms of displacement, velocity, and time. When calculating how long it takes the grenade to travel a horizontal distance of 15.8 meters, kinematics equations come into play. These equations relate the initial velocity, time of flight, and distance traveled.
  • For horizontal motion: The grenade's relative motion combines with the relative velocity of the cars.
  • This combination is used to find how long the grenade has to stay in the air to make the needed distance.
Thus, understanding kinematics allows us to accurately plan the grenade's trajectory and ensure it reaches the intended target.
Velocity Components
When discussing projectile motion, breaking down the initial velocity into its horizontal and vertical components is crucial. These components help solve problems involving angles, like the 45-degree angle in this exercise.
  • The horizontal component (\(v_{0x}\)) is found using the cosine of the angle: \(v_0 \cos(45^\circ)\).
  • The vertical component (\(v_{0y}\)) is found using the sine of the angle: \(v_0 \sin(45^\circ)\).
These components are equal since the angle is 45 degrees, simplifying calculations. The horizontal component determines how far forward the grenade travels, while the vertical component influences how high it goes and how gravity acts on it over time.
Understanding these components allows us to accurately predict and adjust the projectile's path.
Angle of Projection
The angle at which a projectile is launched significantly impacts its trajectory and distance traveled. In our scenario, the hero throws the grenade at a 45-degree angle above the horizontal.

This angle is particularly important because it usually provides the maximum range for a given initial velocity when launching projectiles on level ground. At this angle, the velocity components are equal, providing balanced vertical lift and horizontal coverage. This balance helps the projectile reach the required horizontal distance without compromising on height and descent time.
  • At 45 degrees: both the vertical and horizontal components contribute equally to covering distance and achieving height.
  • This choice of angle simplifies calculations while optimizing the throw's effectiveness.
Knowing how to select the right angle of projection is critical for solving problems involving projectiles and maximizing the effectiveness of the throw.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A canoe has a velocity of 0.40 \(\mathrm{m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing 0.50 \(\mathrm{m} / \mathrm{s}\) east rela- tive to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

The Champion Jumper of the Insect World. The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of 58.7 \(\mathrm{cm}\) above the level ground. (See Nature, Vol. 424 , July \(31,2003,\) p. \(509 . )\) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world- record leap?

Crossing the River I. A river flows due south with a speed of 2.0 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) man steers a motorboat across the river; his velocity relative to the water is 4.2 \(\mathrm{m} / \mathrm{s}\) due east. The river is 800 \mathrm{m}$ wide. (a) What is his velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of his starting point will he reach the opposite bank?

A World Record. In the shot put, a standard track and-field event, a \(7.3-\mathrm{kg}\) object (the shot) is thrown by releasing it at approximately \(40^{\circ}\) over a straight left leg. The world record for distance, set by Randy Barnes in \(1990,\) is 23.11 \(\mathrm{m} .\) Assuming that Barnes released the shot put at \(40.0^{\circ}\) from a height of 2.00 \(\mathrm{m}\) above the ground, with what speed, in \(\mathrm{m} / \mathrm{s}\) and in mph, did he release it?

In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horizontally from the bar (Fig. P3.62). Football regulations are stated in English units, but convert them to SI units for this problem.(a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it is to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) .

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.