Question

For a Science Olympiad competition, a group of middle school students build a trebuchet that can fire a tennis ball from a height of \(1.55 \mathrm{~m}\) with a velocity of \(10.5 \mathrm{~m} / \mathrm{s}\) and a launch angle of \(35.0^{\circ}\) above the horizontal. What is the speed of the tennis ball just before it hits the ground?

Short answer

Answer: The speed of the tennis ball just before it hits the ground is approximately 8.81 m/s.

Step-by-step solution

Identify the given values

We've been given the following values: - Initial height: \(h = 1.55\,\text{m}\) - Initial velocity: \(v_0 = 10.5\,\text{m/s}\) - Launch angle: \(\theta = 35.0^\circ\)

Break down the initial velocity into horizontal and vertical components

The initial horizontal velocity, \(v_{0x}\), and initial vertical velocity, \(v_{0y}\) can be found using the following equations: \(v_{0x} = v_0 \cos \theta\) \(v_{0y} = v_0 \sin \theta\) Substitute the given values and calculate \(v_{0x}\) and \(v_{0y}\): \(v_{0x} = (10.5\,\text{m/s}) \cos 35^\circ \approx 8.62\,\text{m/s}\) \(v_{0y} = (10.5\,\text{m/s}) \sin 35^\circ \approx 6.04\,\text{m/s}\)

Find the time to reach the ground

Now we need to find the time it takes for the tennis ball to hit the ground from that height. We'll use the downward motion equation under constant gravitational acceleration: \(h = v_{0y}t - \frac{1}{2} g t^2\) Where \(g\) is the gravitational acceleration, \(9.81\,\text{m/s}^2\). Rearrange the equation for \(t\): \(t = \frac{2(h - v_{0y}t)}{g}\) Plug in the values and solve for \(t\): \(t \approx 0.84\,\text{s}\)

Determine the final vertical velocity

Now we need to find the final vertical velocity just before the tennis ball hits the ground. We can use the constant acceleration equation: \(v_y = v_{0y} - gt\) Substitute the values and calculate \(v_y\): \(v_y = 6.04\,\text{m/s} - (9.81\,\text{m/s}^2)(0.84\,\text{s}) \approx -2.22\,\text{m/s}\)

Calculate the speed of the tennis ball just before it hits the ground

The final velocity vector before the tennis ball hits the ground will have a horizontal component of \(v_{0x}\) and a vertical component of \(v_y\). We'll now find the magnitude of this vector, which represents the speed of the tennis ball just before it hits the ground: \(v = \sqrt{v_{0x}^2 + v_y^2}\) Substitute the values and calculate \(v\): \(v = \sqrt{(8.62\,\text{m/s})^2 + (-2.22\,\text{m/s})^2} \approx 8.81\,\text{m/s}\) Therefore, the speed of the tennis ball just before it hits the ground is approximately \(\boldsymbol{8.81\,\text{m/s}}\).

Key concepts

Trajectory of Projectiles

Understanding the trajectory of a projectile involves analyzing the path that an object follows when it is thrown or launched into the air. In our exercise, a tennis ball is projected from a trebuchet, and the path it follows is influenced by the initial speed, the angle of launch, and the height from which it is launched.

Imagine the tennis ball as traveling along a curved path; this curve is known as the projectile's trajectory. The shape of this trajectory is parabolic due to the influence of gravity acting on the projectile in a downward direction. Even though the trebuchet givers the tennis ball an initial vertical and horizontal velocity, it's the acceleration due to gravity that changes the ball's motion over time, causing it to eventually fall back to the ground.

Furthermore, the horizontal motion is at a constant speed since there are no horizontal forces acting on the tennis ball (assuming air resistance is negligible), while the vertical motion is affected by gravity, which accelerates the ball downwards.

Kinematic Equations

Kinematic equations are fundamental tools in physics used to describe the motion of an object under constant acceleration. In projectile motion, these equations help us determine various aspects such as the time of flight, range, maximum height, and the speed upon impact. The given problem demonstrates the use of such equations to navigate through the different stages of motion of the tennis ball.

In our exercise improvement advice, we would emphasize checking the signs allotted to the different components of velocities and accelerations. A common mistake is to confuse the direction of the velocity or acceleration vectors, so it is crucial to establish a coordinate system where, typically, upward and forward directions are considered positive, while downward and backward directions are negative.

Breaking Down Initial Velocity

As shown in the step-by-step solution, the initial velocity is broken down into horizontal and vertical components using trigonometry: \(v_{0x} = v_0 \cos \theta\) and \(v_{0y} = v_0 \sin \theta\). These components are used in conjunction with the kinematic equations to understand how the horizontal and vertical motions behave independently.

Gravitational Acceleration

Gravitational acceleration (\(g\) is a universal force that pulls objects toward the Earth's center, with an average value of approximately \(9.81\text{m/s}^2\) on the Earth's surface. It is this acceleration that causes the vertical velocity component of a projectile to change over time, contrary to the horizontal component, which remains constant if air resistance is ignored.

All objects, regardless of their mass, will experience the same gravitational acceleration when falling freely under gravity's influence. This factor is crucial for correctly applying the kinematic equations to the projectile's vertical motion, as demonstrated in the tennis ball example. One of the critical uses of gravitational acceleration in our calculation is to determine the time it takes for the projectile to hit the ground after being launched, using the equation \(h = v_{0y}t - \frac{1}{2} g t^2\) and then further to find the final vertical velocity component just before impact.

An extra tip for accuracy in homework exercises is to always maintain the same units throughout the calculation, typically meters per second for velocity and seconds for time, to obtain the correct motion analysis under the influence of gravity.