Question

A cannon is fired from a hill \(116.7 \mathrm{~m}\) high at an angle of \(22.7^{\circ}\) with respect to the horizontal. If the muzzle velocity is \(36.1 \mathrm{~m} / \mathrm{s},\) what is the speed of a 4.35 -kg cannonball when it hits the ground \(116.7 \mathrm{~m}\) below?

Short answer

Based on the given information (cannon's height: 116.7 m, firing angle: 22.7 degrees, and muzzle velocity: 36.1 m/s), calculate the speed of the 4.35 kg cannonball when it hits the ground: 1. Determine the horizontal and vertical components of the initial velocity. 2. Find the time it takes for the cannonball to hit the ground using the vertical motion equation. 3. Calculate the horizontal distance traveled by the cannonball using the horizontal motion equation. 4. Determine the final vertical velocity and find the final speed. After following these steps, we find that the speed of the 4.35-kg cannonball when it hits the ground, 116.7 meters below, is approximately 47.171 m/s.

Step-by-step solution

Determine the horizontal and vertical components of the initial velocity

Given the muzzle velocity \(v_0\) and the firing angle \(\theta\), we can find the components of the initial velocity using the following trigonometric formulas: Horizontal component: \(v_{0x} = v_0 \cos(\theta)\) Vertical component: \(v_{0y} = v_0 \sin(\theta)\) Now, let's calculate the horizontal and vertical components: \(v_{0x} = 36.1 \cos(22.7^\circ) \approx 32.341 \mathrm{~m/s}\) \(v_{0y} = 36.1 \sin(22.7^\circ) \approx 14.258 \mathrm{~m/s}\)

Find the time it takes for the cannonball to hit the ground

To find the time it takes for the cannonball to hit the ground, we will use the following vertical motion equation: \(h = v_{0y} t - \frac{1}{2} g t^2\) where \(h\) is the vertical displacement (positive upwards), \(v_{0y}\) is the initial vertical velocity, \(t\) is the time, and \(g\) is the gravitational acceleration (approximately 9.81 m/s²). In this case, we are considering the downward displacement of the cannonball (\(h = -116.7 \mathrm{~m}\)). Plugging in the values, we get: \(-116.7 = 14.258 t - 4.905 t^2\) Using a calculator, we can solve this quadratic equation for \(t\) and find the positive value of time: \(t \approx 4.829 \mathrm{~s}\) This is the time it takes for the cannonball to hit the ground.

Calculate the horizontal distance traveled by the cannonball

Using the horizontal motion equation \(x = v_{0x}t\), we can find the horizontal distance traveled by the cannonball: \(x = 32.341 \mathrm{~m/s} \times 4.829 \mathrm{~s} \approx 156.112 \mathrm{~m}\)

Determine the final vertical velocity and find the final speed

At the end of its trajectory, the cannonball's vertical velocity will have changed due to gravitational acceleration. We can find the final vertical velocity using the equation: \(v_y = v_{0y} - gt\) Plugging in the values: \(v_y = 14.258 \mathrm{~m/s} - (9.81 \mathrm{~m/s^2} \times 4.829 \mathrm{~s}) \approx -34.503 \mathrm{~m/s}\) Note that the vertical velocity is negative, indicating that the cannonball is moving downwards when it hits the ground. Finally, let's find the final speed of the cannonball by considering both the horizontal and vertical velocities: \(Speed = \sqrt{v_x^2 + v_y^2}\) \(Speed = \sqrt{(32.341 \mathrm{~m/s})^2 + (-34.503 \mathrm{~m/s})^2} \approx 47.171 \mathrm{~m/s}\) So the speed of the 4.35-kg cannonball when it hits the ground, 116.7 meters below, is approximately \(47.171 \mathrm{~m/s}\).

Key concepts

Muzzle Velocity

Muzzle velocity is the speed at which a projectile, such as a cannonball, exits the barrel of a firearm or the mouth of a cannon.
The muzzle velocity is crucial in projectile motion physics as it directly affects the projectile's range, height, and overall trajectory. When a cannon fires a ball, the muzzle velocity imparts the initial kinetic energy necessary for the projectile to overcome gravitational forces and air resistance.
Understanding muzzle velocity helps to determine the potential distance traveled and impact energy when the projectile lands, which is essential in both military applications and in calculating the physics of ballistics.

Trigonometric Components of Velocity

The trigonometric components of velocity involve breaking down the muzzle velocity into horizontal and vertical components. This is done using basic trigonometry:

• Horizontal component: given by the muzzle velocity times the cosine of the launch angle.
• Vertical component: given by the muzzle velocity times the sine of the launch angle.

These components are essential since they allow us to analyze the projectile's motion separately in the horizontal and vertical directions. The horizontal velocity remains constant (ignoring air resistance), while the vertical velocity changes due to gravitational acceleration. This separation into components is the key to solving many projectile motion problems in physics.

Kinematic Equations

Kinematic equations describe the motion of objects and are vital in understanding projectile motion. They relate the displacement, velocity, acceleration, and time of an object's trajectory.

• The equation to find vertical displacement: \( h = v_{0y} t - \frac{1}{2} g t^2 \), where \( h \) is height, \( v_{0y} \) the initial vertical velocity, \( t \) the time, and \( g \) the gravitational acceleration.
• The equation to find final velocity: \( v = v_0 + a t \), where \( v \) is final velocity, \( v_0 \) initial velocity, \( a \) acceleration, and \( t \) time.

For projectile motion, kinematic equations help determine how long the projectile will remain in the air, how high it will go, and how fast it will be moving when it strikes the ground.

Gravitational Acceleration

Gravitational acceleration is the acceleration due to gravity that acts on objects in freefall near the Earth's surface. Commonly denoted as \( g \), it has an approximate value of \( 9.81 \, m/s^2 \) downwards.
Gravitational acceleration is an essential factor in projectile motion as it constantly acts on the projectile in the vertical direction, affecting the vertical component of its velocity. In kinematic equations, \( g \) is used to calculate the time of flight, the height of the projectile at different points in its motion, and its final speed upon impact. It's a pivotal constant that allows for the prediction and analysis of the paths of projectiles.