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(II) A \(0.40-\mathrm{kg}\) ball is thrown with a speed of \(8.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(36^{\circ} .(a)\) What is its speed at its highest point, and \((b)\) how high does it go? (Use conservation of energy.)

Short Answer

Expert verified
(a) The speed at the highest point is approximately 6.9 m/s. (b) The maximum height is approximately 4.24 m.

Step by step solution

01

Determine the initial vertical and horizontal velocity components

To find the components of the initial velocity, use trigonometry. The initial speed is given as \(8.5 \text{ m/s}\) at an angle of \(36^{\circ}\). The vertical component \(v_{iy}\) is calculated as \(v_{iy} = v_i \sin(\theta) = 8.5 \sin(36^{\circ})\). The horizontal component \(v_{ix}\) is \(v_{ix} = v_i \cos(\theta) = 8.5 \cos(36^{\circ})\).
02

Calculate the speed at the highest point

At the highest point, the vertical component of the velocity \(v_{iy}\) becomes zero because the ball stops ascending before starting to descend. However, the horizontal component \(v_{ix}\) remains constant. Therefore, the speed of the ball at the highest point is equal to its initial horizontal component, \(v_{ix} = 8.5 \cos(36^{\circ})\).
03

Use conservation of energy to find the maximum height

Apply the conservation of energy principle. Initially, the kinetic energy is \(E_{k_i} = \frac{1}{2}mv_i^2\), and at the highest point, all energy is potential, \(E_{p_f} = mgh_{max}\). Setting initial kinetic energy equal to potential energy at the highest point gives \(\frac{1}{2} m v_{iy}^2 = m g h_{max}\). Solve for \(h_{max}\) as \(h_{max} = \frac{v_{iy}^2}{2g}\), where \(v_{iy} = 8.5 \sin(36^{\circ})\).
04

Calculate the maximum height

Substitute the known values: \(v_{iy} = 8.5 \sin(36^{\circ})\) into the equation \(h_{max} = \frac{v_{iy}^2}{2g}\) with \(g = 9.8 \text{ m/s}^2\). This provides the maximum height. Calculate each value step by step to ensure accuracy.

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Key Concepts

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

Velocity Components
When analyzing projectile motion, understanding the velocity components is critical. Velocity is a vector, meaning it has both a magnitude and direction. For a projectile thrown at an angle, it is essential to break down the initial velocity into horizontal and vertical components.

The horizontal component of velocity, denoted as \(v_{ix}\), is calculated using the cosine of the angle of projection. It is given by the formula \(v_{ix} = v_i \cos(\theta)\). The vertical component, \(v_{iy}\), uses the sine of the angle and is calculated as \(v_{iy} = v_i \sin(\theta)\).

These components are crucial in predicting the projectile's trajectory and determining its speed at various points of motion. The horizontal component \(v_{ix}\) remains constant throughout the projectile's flight, unaffected by the force of gravity.
Kinetic Energy
The kinetic energy of an object in motion is given by the formula \(E_k = \frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) is its velocity. In the context of the exercise, when the ball is initially thrown, it has kinetic energy due to its initial velocity.

At different points of the projectile's journey, the velocity changes due to the effect of gravity, thus altering kinetic energy. However, using the principle of conservation of energy, we can set initial kinetic energy equal to potential energy at specific points, like the highest point of its path from which we derived the height it reaches.

This relationship helps us to translate kinetic energy at the beginning into potential energy at the peak of its flight, where the vertical component is zero and the horizontal component maintains the constant kinetic energy for the projectile motion.
Potential Energy
Potential energy, particularly gravitational potential energy, is related to an object's position in a gravitational field. It is given by the equation \(E_p = mgh\), where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is the height above the reference point.

In this exercise, when the ball reaches its highest point, all kinetic energy due to the vertical component of velocity is converted into potential energy. Therefore, at this apex, the formula establishes \(mgh_{max} = \frac{1}{2}mv_{iy}^2\).

This interchange between kinetic and potential energy demonstrates conservation of energy in a closed system. It allows us to determine the maximum height the ball reaches by using information about its initial motion.
Projectile Motion
Projectile motion refers to the path of an object under the influence of gravity that moves along a curved trajectory. It's governed by the kinematic equations of motion and can be broken into horizontal and vertical components.

In projectile motion, the horizontal component of the motion is constant due to the absence of horizontal forces (assuming no air resistance). The vertical component is affected by gravity, causing the object to accelerate downwards. At the top of its arc, the vertical velocity equals zero before gravity pulls it back down.

In our exercise, understanding projectile motion allows us to make predictions about the ball's behavior, such as reaching the highest point where its vertical speed is zero, and how energy conversion helps determine this point. Using conservation of energy within this context clarifies dynamics of projectile motion.

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Most popular questions from this chapter

Suppose the gravitational potential energy of an object of mass \(m\) at a distance \(r\) from the center of the Earth is given by $$ U(r)=-\frac{G M m}{r} e^{-\alpha r} $$ where \(\alpha\) is a positive constant and \(e\) is the exponential function. (Newton's law of universal gravitation has \(\alpha=0\) ). \((a)\) What would be the force on the object as a function of \(r ?(b)\) What would be the object's escape velocity in terms of the Earth's radius \(R_{\mathrm{E}} ?\)

(I) For a satellite of mass \(m_{\mathrm{s}}\) in a circular orbit of radius \(r_{\mathrm{S}}\) around the Earth, determine \((a)\) its kinetic energy \(K,\) (b) its potential energy \(U(U=0\) at infinity \(),\) and \((c)\) the ratio \(K / U\)

(II) A 72 -kg trampoline artist jumps vertically upward from the top of a platform with a \(\begin{array}{llll}\text { speed of } & 4.5 \mathrm{~m} / \mathrm{s} . & (a) & \text { How }\end{array}\) fast is he going as he lands on the trampoline, \(2.0 \mathrm{~m}\) below (Fig. \(8-31\) )? (b) If the trampoline behaves like a spring of spring constant \(5.8 \times 10^{4} \mathrm{~N} / \mathrm{m}\), how far does he depress it?

(II) A spring with \(k=63 \mathrm{~N} / \mathrm{m}\) hangs vertically next to a ruler. The end of the spring is next to the \(15-\mathrm{cm}\) mark on the ruler. If a \(2.5-\mathrm{kg}\) mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

(II) A cyclist intends to cycle up a \(9.50^{\circ}\) hill whose vertical height is \(125 \mathrm{~m}\). The pedals turn in a circle of diameter \(36.0 \mathrm{~cm} .\) Assuming the mass of bicycle plus person is \(75.0 \mathrm{~kg},\) (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves the bike \(5.10 \mathrm{~m}\) along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.

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