/*! 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 14 A ball is thrown vertically up w... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 14

A ball is thrown vertically up with a velocity of \(30 \mathrm{m} / \mathrm{s}\) at the edge of a \(60-\mathrm{m}\) cliff. Calculate the height \(h\) to which the ball rises and the total time \(t\) after release for the ball to reach the bottom of the cliff. Neglect air resistance and take the downward acceleration to be \(9.81 \mathrm{m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The ball rises 45.87 m above the cliff and takes 7.70 s to hit the ground.

Step by step solution

01

Determine the Maximum Height of the Ball

The maximum height is the point where the velocity of the ball becomes zero. Use the kinematic equation for velocity: \[ v^2 = u^2 + 2as \]where \( v = 0 \) (velocity at maximum height), \( u = 30 \mathrm{m/s} \) (initial velocity), and \( a = -9.81 \mathrm{m/s^2} \) (acceleration due to gravity). Substitute these values to find the height above the cliff:\[ 0 = 30^2 + 2(-9.81)h \]Solve for \( h \):\[ h = \frac{30^2}{2 \times 9.81} \]\[ h = \frac{900}{19.62} \]\[ h \approx 45.87 \text{ meters} \].
02

Calculate Total Height from Ground

The total height from the ground to the peak of the ball's trajectory is the sum of the height of the cliff and the height the ball rises above the cliff. So:\[ H = 60 + 45.87 \approx 105.87 \text{ meters} \].
03

Compute Total Time of Flight

To find the total time to reach the bottom of the cliff, calculate the time to reach the maximum height first, then the time to fall from the top of its trajectory to the bottom of the cliff. **Time to reach maximum height:** Use the kinematic equation:\[ v = u + at \]\[ 0 = 30 + (-9.81)t \]Solving for \( t \):\[ t = \frac{30}{9.81} \approx 3.06 \text{ seconds} \].**Time to fall from the maximum height to the ground:** Use the equation:\[ h = \frac{1}{2}gt^2 \], where \( h = 105.87 \text{ meters} \), and \( g = 9.81 \text{ m/s}^2 \). Solving for \(t\):\[ 105.87 = \frac{1}{2} \times 9.81 \times t^2 \]\[ t^2 = \frac{105.87 \times 2}{9.81} \approx 21.59 \]\[ t \approx 4.64 \text{ seconds} \].**Total time:**\[ T = 3.06 + 4.64 = 7.70 \text{ seconds} \].
04

Conclusion: Solution to the Problem

The height the ball rises above the cliff is approximately \( 45.87 \text{ meters} \), making the total height from the base of the cliff \( 105.87 \text{ meters} \). The total time for the ball to hit the ground after release is \( 7.70 \text{ seconds} \).

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Key Concepts

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

Velocity
Velocity is a fundamental concept in kinematics, describing the speed and direction of a moving object. In this exercise, the initial velocity is given as 30 meters per second (m/s) upwards, which means the ball is hurled directly upward off the cliff. The velocity is
  • a vector quantity
  • has both magnitude (size) and direction
  • different from speed, which is scalar and only measures magnitude
Understanding velocity is crucial for predicting the ball's movement and determining the maximum height it reaches. As the ball ascends, the initial positive velocity will decrease to zero at the peak of its trajectory due to the opposing force of gravity. At this point, velocity is zero, marking the transition from ascending to descending motion. Knowing how velocity changes allows us to apply the appropriate formulas to find other parameters of motion, such as time and distance traveled.
Acceleration Due to Gravity
Acceleration due to gravity (often denoted as \( g \)) is a constant force that pulls objects toward the Earth's surface. It is approximately 9.81 meters per second squared (m/s虏), acting in the downward direction. This force is independent of the mass of the objects and impacts any freely falling body. In our problem, the ball is subjected to this force, which slows its upward motion until it stops, and then accelerates it downwards. When solving kinematic problems, acceleration due to gravity can be applied in different equations to find unknowns such as time and distance. Here, negative acceleration is used when the motion is upward because gravitational pull is opposite to the direction of motion, while it is positive when considering the downward journey of the ball. Taking gravity into account is essential for a complete analysis of any projectile's motion.
Trajectory Analysis
Trajectory analysis involves calculating the path that an object follows through space. For the ball thrown from the cliff, its path is influenced by the upward force of the throw and the downward pull of gravity. This projectile path can be visualized and calculated using kinematic equations. In this scenario, the ball rises to a maximum height before descending past the point of launch down to the base of the cliff. The trajectory can be broken into two segments:
  • Ascent to the maximum height: The velocity decreases to zero under the influence of gravity.
  • Descent back to the ground: The velocity increases as gravity pulls it downward.
To analyze the trajectory, two main aspects are considered:
  • Maximum Height: Calculated when the velocity is zero at the top of the trajectory, providing insight into how high above the launch point the projectile reaches.
  • Total Flight Duration: The sum of time taken to ascend to maximum height and the time to descend to the ground.
Understanding these will help you predict and calculate where and when the object will hit the ground, as well as provide insights into how changes in initial conditions affect the motion of projectiles.

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Most popular questions from this chapter

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