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A rocket is fired at a speed of \(75.0 \mathrm{m} / \mathrm{s}\) from ground level, at an angle of \(60.0^{\circ}\) above the horizontal. The rocket is fired toward an \(11.0-\mathrm{m}-\) high wall, which is located \(27.0 \mathrm{m}\) away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Short Answer

Expert verified
The rocket clears the top of the wall by 33.22 meters.

Step by step solution

01

Resolve the Initial Velocity

The initial velocity \(v_0\) of the rocket is given as 75.0 m/s. We need to resolve this velocity into horizontal and vertical components. The horizontal component \(v_{0x}\) is given by \(v_{0x} = v_0 \cdot \cos(\theta)\), and the vertical component \(v_{0y}\) is given by \(v_{0y} = v_0 \cdot \sin(\theta)\), where \(\theta = 60.0^{\circ}\). Calculate these components:\[v_{0x} = 75.0 \cdot \cos(60.0^{\circ}) = 75.0 \cdot 0.5 = 37.5 \mathrm{m/s}\]\[v_{0y} = 75.0 \cdot \sin(60.0^{\circ}) = 75.0 \cdot \frac{\sqrt{3}}{2} \approx 64.95 \mathrm{m/s}\]
02

Calculate Time to Reach the Wall

To find the time \(t\) it takes for the rocket to reach the wall horizontally, we use the formula \(d = v_{0x} \cdot t\), where \(d = 27.0 \mathrm{m}\). Solve for \(t\):\[27.0 = 37.5 \cdot t \Rightarrow t = \frac{27.0}{37.5} \approx 0.72 \mathrm{s}\]
03

Determine Vertical Position at the Wall

Next, calculate the vertical position of the rocket at the time it reaches the wall. The vertical position at any time \(t\) is given by the formula: \(y = v_{0y} \cdot t - \frac{1}{2}gt^2\), where \(g = 9.81 \mathrm{m/s^2}\) is the acceleration due to gravity. Substituting for \(t = 0.72\, \mathrm{s}\):\[y = 64.95 \cdot 0.72 - \frac{1}{2} \cdot 9.81 \cdot (0.72)^2\]Calculate \(y\):\[y = 46.764 - 2.54 \approx 44.22 \mathrm{m}\]
04

Calculate the Clearance Over the Wall

The wall is 11.0 m high. The clearance \(C\) over the wall is the difference between the rocket's height at \(t = 0.72\, \mathrm{s}\) and the height of the wall. Compute \(C\):\[C = 44.22 - 11.0 = 33.22 \mathrm{m}\]

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Key Concepts

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

Kinematics in Projectile Motion
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. When we talk about projectile motion, we're looking at the path an object takes when it's launched into the air and is subject to gravity. This type of motion can be broken down into two components: horizontal and vertical. By analyzing these components separately, we can more easily predict where an object will be at any given time during its flight. This separation is crucial for solving problems related to projectiles, such as the one with the rocket in the exercise. Understanding how an initial velocity is divided into horizontal and vertical parts lays the groundwork for understanding the entire motion.
Horizontal and Vertical Components
Each projectile's motion can be divided into horizontal and vertical components, which are calculated using trigonometric functions. The horizontal component (\(v_{0x}\)) of a projectile's velocity is calculated with the cosine of the launch angle, while the vertical component (\(v_{0y}\)) is determined using the sine. These calculations allow us to analyze the projectile's trajectory independently in both directions.
  • The horizontal motion of a projectile is constant because gravity does not affect it.
  • The vertical motion, however, is influenced by gravity, which causes the projectile to slow down as it ascends and speed up as it descends.
For the given rocket example, resolving the initial velocity was crucial to determine where it would be at each point along its flight path.
Acceleration Due to Gravity
Gravity is the force that constantly acts on a projectile, influencing its vertical motion. It is always directed downward and is approximately \(9.81 \, \text{m/s}^2\) on Earth. This acceleration influences the vertical velocity of the projectile, decelerating it as it moves upwards and accelerating it downwards after it reaches its peak.
Understanding this concept helps us calculate how high an object will ascend before gravity causes it to fall back to the ground. In the case of the rocket, once it reaches its highest point, it will begin to descend due to gravity. To find its position at any time, the formula \(y = v_{0y} \cdot t - \frac{1}{2}gt^2\) was used, taking into consideration that the horizontal movement is unaffected by gravity.
Trigonometric Functions in Projectile Motion
Trigonometric functions like sine and cosine are essential tools in breaking down a vector, like the rocket's initial velocity, into horizontal and vertical components. These functions relate the angles of a right triangle to the lengths of its sides, allowing us to calculate these components when given an angle and an initial velocity.
  • The cosine of the angle gives the ratio of the adjacent side over the hypotenuse, which helps determine the horizontal component.
  • The sine of the angle provides the ratio of the opposite side over the hypotenuse, used to find the vertical component.
By applying these trigonometric functions, the exercise resolved the rocket's velocity into two parts, which is a fundamental step in solving any projectile motion problem.

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Most popular questions from this chapter

Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown horizontally with a speed of \(15 \mathrm{m} / \mathrm{s}\) from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of \(65^{\circ}\) with the horizontal. From what height above the ground was the marble thrown?

The earth moves around the sun in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{m} .\) During the three summer months (an elapsed time of \(\left.7.89 \times 10^{6} \mathrm{s}\right),\) the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?

A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at a distance of \(H_{A}\) below the center, however. The second time, the rifle is similarly aimed, but from twice the distance from the target. This time the bullet strikes the target at a distance of \(H_{B}\) below the center. Find the ratio \(H_{B} / H_{A}\)

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be \(162 \mathrm{km}\) away. The radar antenna is pointing upward at an angle of \(62.3^{\circ}\) from the ground. Find the \(x\) and \(y\) components (in \(\mathrm{km}\) ) of the position vector of the satellite, relative to the antenna.

A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{m} / \mathrm{s}\), due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

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