Question

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

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

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.

Convert the angle from degrees to radians

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

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): $$ t = \frac{v_{0y}}{g} $$

Calculate the maximum height above the ground

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

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.