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Chapter 6: Problem 38

A cannonball of mass \(5.99 \mathrm{~kg}\) is shot from a cannon at an angle of \(50.21^{\circ}\) relative to the horizontal and with an initial speed of \(52.61 \mathrm{~m} / \mathrm{s}\). As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?

Short Answer

Expert verified
Answer: To calculate the gain in potential energy, follow these steps: 1. Calculate the vertical component of the initial velocity using the formula: \(v_{0y} = v_0 \times \sin(\theta)\). 2. Convert the angle from degrees to radians using the formula: \(\theta(rad) = \frac{\theta(deg) \times \pi}{180}\). 3. Calculate the time taken to reach the highest point using the formula: \(t = \frac{v_{0y}}{g}\). 4. Calculate the maximum height above the ground using the formula: \(h = v_{0y}t - \frac{1}{2}gt^2\). 5. Calculate the potential energy gained using the formula: \(\Delta PE = mgh\). By following these steps, you will find the gain in potential energy at the highest point of the cannonball's trajectory.

Step by step solution

01

Calculate the vertical component of the initial velocity

Using the angle of projection (50.21°) and the initial speed (52.61 m/s), we can determine the initial vertical velocity component. We use the trigonometric sine function: $$ v_{0y} = v_0 \times \sin(\theta) $$ where \(v_0\) is the initial speed (52.61 m/s), \(\theta\) is the angle (50.21°), and \(v_{0y}\) is the vertical component of the velocity.
02

Convert the angle from degrees to radians

To use the sine function, we need to convert the angle from degrees to radians: $$ \theta(rad) = \frac{\theta(deg) \times \pi}{180} $$
03

Calculate the time taken to reach the highest point

The time taken to reach the highest point can be calculated using the kinematic formula: $$ v_y = v_{0y} - gt $$ At the highest point, the vertical velocity (\(v_y\)) will be 0. So solving for \(t\), where \(g\) is the gravitational acceleration (9.81 m/s²): $$ t = \frac{v_{0y}}{g} $$
04

Calculate the maximum height above the ground

Use the time calculated above to find the maximum height reached by the cannonball: $$ h = v_{0y}t - \frac{1}{2}gt^2 $$
05

Calculate the potential energy gained

The gain in potential energy is found by multiplying the mass, gravitational acceleration, and the maximum height: $$ \Delta PE = mgh $$ Where \(m\) is the mass of the cannonball (5.99 kg), \(g\) is gravitational acceleration (9.81 m/s²), and \(h\) is the maximum height.

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Most popular questions from this chapter

A father exerts a \(2.40 \cdot 10^{2} \mathrm{~N}\) force to pull a sled with his daughter on it (combined mass of \(85.0 \mathrm{~kg}\) ) across a horizontal surface. The rope with which he pulls the sled makes an angle of \(20.0^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.200,\) and the sled moves a distance of \(8.00 \mathrm{~m}\). Find a) the work done by the father, b) the work done by the friction force, and c) the total work done by all the forces.

A \(1.50 \cdot 10^{3}-\mathrm{kg}\) car travels \(2.50 \mathrm{~km}\) up an incline at constant velocity. The incline has an angle of \(3.00^{\circ}\) with respect to the horizontal. What is the change in the car's potential energy? What is the net work done on the car?

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)

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.
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