/*! 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 9 . Fast Bullets A rifle that shoo... [FREE SOLUTION] | 91Ó°ÊÓ

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. Fast Bullets A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away and level with the rifle. How high above the target must the rifle barrel be pointed so that the bullet hits the target?

Short Answer

Expert verified
The rifle must be aimed 0.048 meters above the target.

Step by step solution

01

- Determine Time of Flight

Use the formula for constant velocity to find the time of flight. The formula is given by \[ t = \frac{d}{v} \] where \(d = 45.7 \text{ m}\) and \(v = 460 \text{ m/s} \). Substitute the known values to get \[ t = \frac{45.7 \text{ m}}{460 \text{ m/s}} \approx 0.099 \text{ seconds} \].
02

- Calculate Vertical Drop

The bullet is subjected to gravity during its flight. Use the formula for the vertical displacement under constant acceleration: \[ y = \frac{1}{2}gt^2 \] where \(g \approx 9.8 \text{ m/s}^2\) and \(t = 0.099 \text{ seconds}\). Substitute the values to get \[ y = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (0.099 \text{ s})^2 \approx 0.048 \text{ m} \].
03

- Determine Height Above Target

The height above the target that the rifle barrel must be aimed is equal to the vertical drop calculated in step 2. Therefore, the rifle must be aimed \( 0.048 \text{ m}\) above the target.

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Key Concepts

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

Time of Flight
To understand how far a bullet will travel horizontally before it hits the target, we need to calculate its time of flight. The bullet travels horizontally with a constant velocity given by the rifle's muzzle speed, which in this scenario is 460 m/s. The formula to find the time of flight is \[ t = \frac{d}{v} \] where \(t\) is the time, \(d\) is the distance to the target, and \(v\) is the muzzle speed of the bullet. For the given problem, with the target at 45.7 meters away, and the speed of 460 meters per second, we substitute these values into the formula: \[ t = \frac{45.7 \text{ m}}{460 \text{ m/s}} \approx 0.099 \text{ seconds} \] This means the bullet will be in the air for about 0.099 seconds before reaching the target.
Vertical Displacement
While the bullet is traveling towards the target, gravity pulls it downwards. This downward movement is known as vertical displacement. We can calculate the vertical drop by using the formula for vertical displacement under constant acceleration, which in this case is due to gravity: \[ y = \frac{1}{2}gt^2 \] Here, \(g\) represents the acceleration due to gravity (approximately 9.8 m/s²), and \(t\) is the time of flight, which we previously calculated as 0.099 seconds. Substituting these values in, we get: \[ y = \frac{1}{2} \times 9.8 \text{ m/s}^2 \times (0.099 \text{ s})^2 \approx 0.048 \text{ m} \] Thus, during its flight, the bullet will fall approximately 0.048 meters (or 4.8 centimeters) due to gravity.
Constant Velocity
The concept of constant velocity is crucial in calculating the time of flight for projectiles. When an object moves at constant velocity, it covers equal distances in equal intervals of time, without any change in speed or direction. In our specific problem, the bullet exits the rifle with a constant velocity of 460 meters per second. This means there is no horizontal acceleration acting on the bullet once it exits the rifle. Because the velocity is constant, we can use the simple formula for distance traveled at constant velocity: \[ d = vt \] This relationship was used in the initial step to simplify the calculation of the bullet's time of flight. By understanding that the bullet's horizontal motion remains uniform (unaffected by forces speaking horizontally), we focus solely on vertical factors, such as gravity, to determine other elements like vertical displacement.

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Most popular questions from this chapter

\(A\) rifle is aimed horizontally at a target \(30 \mathrm{~m}\) away. The bullet hits the target \(1.9 \mathrm{~cm}\) below the aiming point. What are (a) the bullet's time of flight and (b) its speed as it emerges from the rifle?

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m}\). (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 \mathrm{~g}\) ? (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?

A boy whirls a stone in a horizontal circle of radius \(1.5 \mathrm{~m}\) and at height \(2.0 \mathrm{~m}\) above level ground. The string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of \(10 \mathrm{~m}\). What is the magnitude of the centripetal acceleration of the stone while in circular motion?

Mike Powell In the 1991 World Track and Field Championships in Tokyo, Mike Powell (Fig. 5-30) jumped \(8.95 \mathrm{~m}\), breaking the 23 -year long-jump record set by Bob Beamon by a full \(5 \mathrm{~cm}\). Assume that Powell's speed on takeoff was \(9.5 \mathrm{~m} / \mathrm{s}\) (about equal to that of a sprinter) and that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\) in Tokyo. How much less was Powell's horizontal range than the maximum possible horizontal range (neglecting the effects of air) for a particle launched at the same speed of \(9.5 \mathrm{~m} / \mathrm{s}\) ?

(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would the period of rotation of Earth have to be for objects on the equator to have a centripetal acceleration with a magnitude of \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

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