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Chapter 6: Problem 41

Suppose you throw a 0.052 -kg ball with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(30.0^{\circ}\) above the horizontal from a building \(12.0 \mathrm{~m}\) high. a) What will be its kinetic energy when it hits the ground? b) What will be its speed when it hits the ground?

Short Answer

Expert verified
Question: A ball of mass 0.052 kg is thrown off a building with an initial speed of 10.0 m/s at an angle of 30 degrees above the horizontal. The building is 12.0 m tall. Calculate a) the speed of the ball when it hits the ground and b) the kinetic energy of the ball when it hits the ground. Answer: Follow the steps outlined in the solution to calculate the initial velocity components, time of flight, horizontal and vertical velocities at impact, speed when the ball hits the ground, and the kinetic energy at impact. Remember to use the given mass, initial speed, launch angle, and height of the building in your calculations.

Step by step solution

01

Determine the initial velocity components

First, we need to find the initial horizontal and vertical components of velocity. We can do that using the given initial speed (10.0 m/s) and launch angle (30 degrees). Using trigonometry and keeping in mind that cosine is for horizontal and sine is for the vertical component: \(v_{0x} = v_{0} \cos(30^{\circ}) = 10.0 \cdot \cos(30^{\circ})\) \(v_{0y} = v_{0} \sin(30^{\circ}) = 10.0 \cdot \sin(30^{\circ})\)
02

Calculate the time of flight

To find the time of flight, we can use the vertical motion equation: \(h = v_{0y}t - \frac{1}{2}gt^2\) where \(h\) is the vertical displacement (12.0 m), \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (9.81 m/s^2), and \(t\) is the time. In this case, the ball is thrown from a height, so \(h\) will be negative (-12.0 m). Rearrange the equation to solve for \(t\).
03

Calculate the horizontal and vertical velocities at impact

Now, use the horizontal velocity (\(v_{0x}\)) and the time of flight (\(t\)) to determine the horizontal displacement. As the horizontal velocity remains constant throughout the motion, the horizontal velocity is the same as the initial horizontal velocity (\(v_{0x}\)). To find the vertical velocity (\(v_y\)) at impact, it's necessary to consider the gravitational acceleration during the time of flight. We can use the vertical component of the kinematic equation for the final velocity: \(v_{y} = v_{0y} - gt\)
04

Calculate the speed when the ball hits the ground

With both the horizontal and vertical velocities at impact, we can find the speed of the ball by using the Pythagorean theorem: \(v = \sqrt{v_{x}^2 + v_{y}^2}\)
05

Calculate the kinetic energy at impact

Finally, we can calculate the kinetic energy of the ball at the moment of impact using the formula for kinetic energy: \(K.E. = \frac{1}{2}mv^2\) where \(m\) is the mass of the ball (0.052 kg) and \(v\) is the speed at impact. After plugging in the calculated values, we'll have the answers for both parts a) and b) of the exercise.

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Most popular questions from this chapter

The greenskeepers of golf courses use a stimpmeter to determine how "fast" their greens are. A stimpmeter is a straight aluminum bar with a V-shaped groove on which a golf ball can roll. It is designed to release the golf ball once the angle of the bar with the ground reaches a value of \(\theta=20.0^{\circ} .\) The golf ball \((\) mass \(=1.62 \mathrm{oz}=0.0459 \mathrm{~kg})\) rolls 30.0 in down the bar and then continues to roll along the green for several feet. This distance is called the "reading." The test is done on a level part of the green, and stimpmeter readings between 7 and \(12 \mathrm{ft}\) are considered acceptable. For a stimpmeter reading of \(11.1 \mathrm{ft},\) what is the coefficient of friction between the ball and the green? (The ball is rolling and not sliding, as we usually assume when considering friction, but this does not change the result in this case.)

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