Question

To determine the gravitational acceleration at the surface of a newly discovered planet, scientists perform a projectile motion experiment. They launch a small model rocket at an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) and an angle of \(30.0^{\circ}\) above the horizontal and measure the (horizontal) range on flat ground to be \(2165 \mathrm{~m}\). Determine the value of \(g\) for the planet.

Short answer

Answer: The gravitational acceleration (g) on the planet is 0.50 m/s².

Step-by-step solution

Decompose the initial velocity into horizontal and vertical components

To analyze the projectile motion, we first need to find the horizontal (vx) and vertical (vy) components of the initial velocity. Using basic trigonometry and the given angle, we can calculate these components as follows: \[v_x = v_0 \cos \theta \] \[v_y = v_0 \sin \theta \] Using the given values, \(v_0 = 50.0 \mathrm{~m/s}\), and \(\theta = 30.0^\circ\), we can find those components: \[v_x = 50.0 \cos (30.0^\circ) = 43.3 \mathrm{~m/s}\] \[v_y = 50.0 \sin (30.0^\circ) = 25.0 \mathrm{~m/s}\]

Determine the time of flight using the horizontal range

To find the time of flight, we will use the horizontal motion, as it is not affected by gravitational acceleration. Since the horizontal velocity is constant, we can use the linear relationship between horizontal range (R) and time (t): \[R = v_x t \] We are given R = 2165 m, and we found \(v_x = 43.3 \mathrm{~m/s}\), so we can solve for t: \[ t = \frac{R}{v_x} = \frac{2165}{43.3} = 50 \mathrm{~s} \]

Use the vertical motion equation to solve for gravitational acceleration (g)

For vertical motion, we use the following kinematic equation: \[v_y^2 = v_{0y}^2 - 2g(y - y_0)\] Since we're considering only the maximum height and launch position, the final and initial vertical positions are equal (\(y = y_0\)). So the equation becomes: \[v_y^2 = v_{0y}^2 - 2gy\] \[0 = v_{0y}^2 - 2gy\] \[g = \frac{v_{0y}^2}{2y}\] The rocket reaches its maximum height when \(v_y = 0\). We can use the first vertical motion equation to find the maximum height (y): \[v_y = v_{0y} - gt \] \[0 = 25.0 - g(50) \] \[g(50) = 25.0 \] \[g = \frac{25.0}{50} = 0.50 \mathrm{~m/s^2} \] So the gravitational acceleration (g) of the planet is \(0.50 \mathrm{~m/s^2}\).

Key concepts

Projectile Motion

Projectile motion involves objects launched into the air and affected by gravity, moving along a curved path. Notably, this motion can be analyzed by decomposing it into horizontal and vertical components that are independent of each other.

Gravity only affects the vertical motion, accelerating the projectile downwards, while the horizontal motion continues at a constant velocity, unaffected by gravity. Understanding this separation allows us to solve complex problems by breaking them into simpler, one-dimensional motions.

For a projectile launched at an angle, the initial velocity contributes to both the vertical lift and the horizontal range. Changes in launch angle or initial speed can create various paths, from steep arcs to long, flat trajectories. The key parameters calculated in projectile motion include the range, maximum height, time of flight, and the final position upon landing.

Initial Velocity Components

When tackling projectile motion, it's crucial to start by breaking down the initial speed into its horizontal (vx) and vertical (vy) components. These components dictate the initial direction and speed of the projectile's motion along each axis.

The horizontal component, calculated as \( v_x = v_0 \cos \theta \) is always constant, because it is not influenced by gravity. On the other hand, the vertical component, calculated as \( v_y = v_0 \sin \theta \), is directly affected by the force of gravity, and changes over time as the projectile ascends and descends.

The initial velocity components are foundational for predicting the projectile's behavior, determining its range, maximum height, and impact point.

Time of Flight

The time of flight refers to the total duration a projectile stays in the air. For horizontal launch scenarios, it only depends on the vertical motion, as the horizontal velocity remains constant.

To calculate the time of flight when the landing height is equal to the launch height, you can use the concept of symmetry in projectile motion. The time to reach the maximum height is equal to the time taken to fall back down from that height. This time is derived from the initial vertical velocity and the gravitational acceleration.

The experiment illustrates this concept by measuring the horizontal range and using the constant horizontal velocity to calculate the time the rocket model is airborne. Devising experiments that manipulate the time a projectile remains in flight can yield insights into the effects of gravitational acceleration and air resistance.

Kinematic Equations

Kinematic equations describe the motion of objects in terms of their velocity, acceleration, time, and displacement. In the context of projectile motion, these equations are indispensable for determining the vertical and horizontal positions and velocities at any point in time.

The equations are built around the assumption of constant acceleration, which fits the case of an object moving under the influence of gravity. Although a projectile’s horizontal velocity is constant (no horizontal acceleration), its vertical velocity changes because of the vertical acceleration due to gravity.

In the provided example, kinematic equations allow us to relate the time of flight to the horizontal range and connect the vertical motion’s initial velocity and displacement to the gravitational acceleration. The application of these formulas yields the gravitational acceleration value on the new planet.