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Chapter 3: Problem 74

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
The grenade's initial velocity relative to the hero is 7.86 m/s, and relative to the earth is 31.09 m/s.

Step by step solution

01

Break Down the Initial Velocities

Convert the speed of both cars from km/h to m/s first.- Hero's car speed: \[ 90.0 \text{ km/h} = \frac{90.0 \times 1000}{3600} \text{ m/s} = 25.0 \text{ m/s} \]- Enemy's car speed: \[ 110 \text{ km/h} = \frac{110 \times 1000}{3600} \text{ m/s} = 30.56 \text{ m/s} \]Calculate the relative velocity of the enemy's car to the hero's car: \[ v_{relative} = v_{enemy} - v_{hero} = 30.56 \text{ m/s} - 25.0 \text{ m/s} = 5.56 \text{ m/s} \]
02

Determine Time of Collision

Use the relative velocity of the enemy's car with respect to the hero's car to calculate the time until the grenade must reach the enemy's car.- Given: Distance to enemy's car is 15.8 mThe time \(t\) can be calculated using the formula: \[ t = \frac{\text{Distance}}{v_{relative}} = \frac{15.8}{5.56} = 2.84 \text{ seconds} \]
03

Resolve Grenade's Initial Velocity Components

Decompose the grenade's initial velocity \(v_i\) into horizontal and vertical components relative to the hero:- Horizontal component: \[ v_{i, x} = v_i \cos(45^{\circ}) = \frac{v_i}{\sqrt{2}} \]- Vertical component: \[ v_{i, y} = v_i \sin(45^{\circ}) = \frac{v_i}{\sqrt{2}} \]
04

Use Kinematic Equation for Horizontal Motion

Since the horizontal motion must account for the relative velocity ensuring the grenade matches the additional speed to reach the enemy's car, we can set the horizontal component equal to the relative velocity: \[ v_{i, x} = v_{relative} = 5.56 \text{ m/s} \]Solving for \(v_i\): \[ \frac{v_i}{\sqrt{2}} = 5.56 \] \[ v_i = 5.56 \sqrt{2} = 7.86 \text{ m/s} \]
05

Calculate Velocity of Grenade Relative to Earth

The earth-bound horizontal velocity of the grenade is the summation of the hero's car speed and the horizontal component of the grenade's velocity:- Earth horizontal velocity: \[ 25.0 + \frac{7.86}{\sqrt{2}} = 25.0 + 5.56 = 30.56 \text{ m/s} \]Since the initial throw is upwards, the vertical component remains the same; hence the earth relative velocity vector is: \[ v_{earth} = \sqrt{(30.56)^2 + (5.56)^2} = 31.09 \text{ m/s} \]

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Key Concepts

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

Relative Velocity
Relative velocity helps us understand how fast one object is moving with respect to another. Imagine two cars moving in the same direction. The speed of one car relative to the other is simply the difference in their velocities. In this exercise, the hero's car moves at 25.0 m/s, and the enemy's car moves at 30.56 m/s - both moving in the same direction.
So, the relative velocity, which is the speed at which the enemy's car seems to approach from the hero's perspective, is calculated as follows:
  • Relative velocity formula: \( v_{relative} = v_{enemy} - v_{hero} \)
  • Substitution: \( v_{relative} = 30.56 \text{ m/s} - 25.0 \text{ m/s} = 5.56 \text{ m/s} \)
This means the enemy's car is moving at 5.56 m/s faster than the hero's car, from the viewpoint of the hero.
Kinematic Equations
Kinematic equations allow us to analyze motion, detailing how an object's position and velocity change over time. These equations are invaluable tools when working with projectile motion, as they help us predict the path and speed of an object.
In this scenario, the relevant kinematic equation helps us find the time during which the grenade reaches the enemy's car. Knowing the relative velocity and distance, we can use the following formula:
  • Time formula: \( t = \frac{\text{Distance}}{v_{relative}} \)
  • Calculation: \( t = \frac{15.8 \text{ m}}{5.56 \text{ m/s}} = 2.84 \text{ seconds} \)
This calculation shows that the grenade must reach the enemy's car within 2.84 seconds to ensure the hero hits his target.
Velocity Components
To accurately describe a projectile like the grenade in this exercise, breaking its velocity into components is crucial. In projectile motion, you split the initial velocity into horizontal and vertical components. This helps in comprehending different influences like gravity acting vertically.
For our grenade, which is launched at an angle of 45 degrees, the initial velocity components are:
  • Horizontal component: \( v_{i, x} = v_i \cos(45^{\circ}) = \frac{v_i}{\sqrt{2}} \)
  • Vertical component: \( v_{i, y} = v_i \sin(45^{\circ}) = \frac{v_i}{\sqrt{2}} \)
In this problem, these components must account for the relative velocity needed for the grenade to catch up with the enemy's car. The equation for the horizontal component is set equal to the relative velocity, helping us solve for \( v_i \), the grenade's initial speed. Solving this gives:
  • \( \frac{v_i}{\sqrt{2}} = 5.56 \)
  • \( v_i = 5.56 \sqrt{2} = 7.86 \text{ m/s} \)
Thus, understanding velocity components helps determine both crucial initial parameters and predict the motion's outcome.

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Most popular questions from this chapter

Leaping the River I. A car comes to a bridge during a storm and finds the bridge washed out The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.8 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?

An airplane pilot sets a compass course due west and maintains an airspeed of 220 \(\mathrm{km} / \mathrm{h}\) . After flying for 0.500 \(\mathrm{h}\) , she finds herself over a town 120 \(\mathrm{km}\) west and 20 \(\mathrm{km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is 40 \(\mathrm{km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 220 \(\mathrm{km} / \mathrm{h}\) .

Centrifuge on Mercury. A laboratory centrifuge on earth makes \(n\) rpm \((\operatorname{rev} / \min )\) and produces an acceleration of 5.00\(g\) at its outer end. (a) What is the acceleration (in \(g^{\prime}\) s) at a point halfway out to the end? (b) This centrifuge is now used in a space capsule on the planet Mercury, where \(g_{\text { Mexary }}\) is 0.378 what it is on earth. How many \(\operatorname{rom}\) (in terms of \(n )\) should it make to produce 5g at its outer end?

A rock is thrown from the roof of a building with a velocity \(v_{0}\) at an angle of \(\alpha_{0}\) from the horizontal. The building has height \(h\) You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_{0}\) -

A rininoceros is at the origin of coordinates at time \(t_{1}=0 .\) For the time interval from \(t_{1}=0\) to \(t_{2}=12.0 \mathrm{s}\) , the rhino's average velocity has \(x-\) component \(-3.8 \mathrm{m} / \mathrm{s}\) and \(y\) -component 4.9 \(\mathrm{m} / \mathrm{s}\) . At time \(t_{2}=12.0 \mathrm{s},(\mathrm{a})\) what are the \(x\) - and \(y\) -coordinates of the rhino? (b) How far is the rhino from the origin?
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