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Chapter 6: Problem 37

A ball is thrown up in the air, reaching a height of \(5.00 \mathrm{~m}\). Using energy conservation considerations, determine its initial speed.

Short Answer

Expert verified
Answer: The initial speed of the ball when it was thrown is approximately 9.95 m/s.

Step by step solution

01

Identifying the given information

We are given the maximum height reached by the ball, which is \(5.00 \mathrm{~m}\). Our task is to determine its initial speed when it was thrown.
02

Understanding the conservation of mechanical energy

The conservation of mechanical energy states that the total mechanical energy (potential + kinetic) in a closed system remains constant. In this case, we'll consider the ball's mechanical energy just when it is thrown and when it reaches its maximum height.
03

Applying the conservation of mechanical energy

When the ball is thrown, it has an initial kinetic energy and zero potential energy. At the maximum height, its potential energy is at the maximum, and its kinetic energy is zero since its velocity is zero. We can write the equation for the conservation of mechanical energy as: \(K_i + U_i = K_f + U_f\) where \(K_i\) - Initial kinetic energy \(U_i\) - Initial potential energy \(K_f\) - Final kinetic energy \(U_f\) - Final potential energy Since the initial potential energy is zero (let's consider the ground level as our reference), and the final kinetic energy is zero (at max height, the ball isn't moving), the equation becomes: \(K_i = U_f\)
04

Calculating the initial and final energies

The formula for the kinetic energy of an object is: \(K = \frac{1}{2}mv^2\) where \(m\) is the mass and \(v\) is the velocity. The formula for the gravitational potential energy is: \(U = mgh\) where \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)) and \(h\) is the height. Now, according to our energy conservation equation, we have: \(\frac{1}{2}mv^2 = mgh\)
05

Solving for the initial speed

We can cancel the mass \(m\) from both sides since it is a common factor: \(\frac{1}{2}v^2 = gh\) Now, we can solve for the initial speed (\(v\)) by isolating it: \(v^2 = 2gh\) \(v = \sqrt{2gh}\) Substitute the given height (\(h = 5.00\mathrm{~m}\)) and acceleration due to gravity (\(g = 9.81 \mathrm{~m/s^2}\)): \(v = \sqrt{(2)(9.81 \mathrm{~m/s^2})(5.00\mathrm{~m})}\) \(v \approx 9.95 \mathrm{~m/s}\) So, the ball's initial speed when it was thrown is approximately \(9.95 \mathrm{~m/s}\).

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Most popular questions from this chapter

A mass of \(1.00 \mathrm{~kg}\) attached to a spring with a spring constant of \(100 .\) N/m oscillates horizontally on a smooth frictionless table with an amplitude of \(0.500 \mathrm{~m} .\) When the mass is \(0.250 \mathrm{~m}\) away from equilibrium, determine: a) its total mechanical energy; b) the system's potential energy and the mass's kinetic energy; c) the mass's kinetic energy when it is at the equilibrium point. d) Suppose there was friction between the mass and the table so that the amplitude was cut in half after some time. By what factor has the mass's maximum kinetic energy changed? e) By what factor has the maximum potential energy changed?

Which of the following is not a unit of energy? a) newton-meter b) joule c) kilowatt-hour d) \(\operatorname{kg} \mathrm{m}^{2} / \mathrm{s}^{2}\) e) all of the above

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?

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

What is the gravitational potential energy of a \(2.0-\mathrm{kg}\) book \(1.5 \mathrm{~m}\) above the floor?
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