Problem 37 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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: \(K_i = U_f\)

Step 4: 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: \(\frac{1}{2}mv^2 = mgh\)

Step 5: 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 \(9.95 \mathrm{~m/s}\).