Problem 44 - College Physics

Question

\bulletA test rocket is launched by accelerating it along a \(200.0-\mathrm{m}\) incline at 1.25 \(\mathrm{m} / \mathrm{s}^{2}\) starting from rest at point \(A\) (Figure \(3.40 . )\) The incline rises at \(35.0^{\circ}\) above the horizontal,and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point \(A .\)

Step-by-Step Solution

Answer

(a) The maximum height above the ground is 123.08 m. (b) The greatest horizontal range beyond point A is 211.77 m.

Step 1: Compute Velocity at the End of the Incline

To find the velocity of the rocket at the end of the incline, use the kinematic equation \( v^2 = u^2 + 2a s \). Here, \( u = 0 \) (initial velocity), \( a = 1.25 \, \mathrm{m/s^2} \) (acceleration), and \( s = 200.0 \, \mathrm{m} \) (distance along the incline).\[v^2 = 0 + 2 \cdot 1.25 \cdot 200 = 500\]\[v = \sqrt{500} = 22.36 \, \mathrm{m/s}\]

Step 2: Decompose Velocity into Components

Calculate the horizontal and vertical components of the velocity at the end of the incline using:\( v_{x} = v \cos(\theta) \) and \( v_{y} = v \sin(\theta) \), where \( \theta = 35.0^{\circ} \).\[v_{x} = 22.36 \cos(35^{\circ}) \approx 18.31 \, \mathrm{m/s}\]\[v_{y} = 22.36 \sin(35^{\circ}) \approx 12.81 \, \mathrm{m/s}\]

Step 3: Determine Maximum Height Above Incline

Use the equation \( v_y^2 = u_y^2 + 2a_sy \) with \( v_y = 0 \) (at maximum height), \( u_y = 12.81 \, \mathrm{m/s} \) (vertical velocity component), and \( a = -9.8 \, \mathrm{m/s^2} \) (acceleration due to gravity) to find the additional height \( h \) above the incline.\[0 = (12.81)^2 + 2(-9.8)h\]\[h = \frac{(12.81)^2}{2 \cdot 9.8} \approx 8.37 \, \mathrm{m}\]

Step 4: Calculate Total Height Above Ground

Add the height of the incline to the maximum height above the incline. The height of the incline is given by \( 200 \sin(35^{\circ}) \).\[h_{\text{incline}} = 200 \sin(35^{\circ}) \approx 114.71 \, \mathrm{m}\]\[h_{\text{total}} = 114.71 + 8.37 = 123.08 \, \mathrm{m}\]

Step 5: Calculate Time of Flight

The total flight time can be found from the vertical motion using the formula \( v_y = u_y + at \) twice, first to the peak and then doubling it (since time up equals time down). At peak, \( v_y = 0 \).\[0 = 12.81 - 9.8t_{\text{up}}\]\[t_{\text{up}} = \frac{12.81}{9.8} \approx 1.31 \, \mathrm{seconds}\]Double this time for total time of flight after leaving the incline.\[t_{\text{total}} = 2 \cdot 1.31 = 2.62 \, \mathrm{seconds}\]

Step 6: Compute Horizontal Range Beyond the Incline

Use the horizontal component of the velocity and total time of flight to find the horizontal range with \( R = v_x \times t_{\text{total}} \).\[R = 18.31 \times 2.62 \approx 47.95 \, \mathrm{m}\]

Step 7: Total Horizontal Distance from A

Add the horizontal component of the incline to the range beyond the incline (since the total distance from A includes the entire incline's length projection on the horizontal).\[R_{\text{incline}} = 200 \cos(35^{\circ}) \approx 163.82 \, \mathrm{m}\]\[R_{\text{total}} = 163.82 + 47.95 = 211.77 \, \mathrm{m}\]

Key Concepts

Kinematics

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause this motion. It uses several equations to relate different parameters of motion, such as velocity, acceleration, displacement, and time. In the given problem, we use the kinematic equation \( v^2 = u^2 + 2as \) to determine the velocity of a rocket at the end of an incline.
This equation relates the initial and final velocities, acceleration, and the distance covered.
In the context of a rocket launch, understanding kinematics helps to predict how the rocket moves along the inclined path, emphasizing time and velocity changes.

• **Initial Velocity ( u )**: In rocket launches starting from rest, initial velocity is zero.
• **Acceleration ( a )**: Uniform acceleration results from the rocket engine’s thrust.
• **Distance ( s )**: The length of the inclined plane determines the distance traveled under constant acceleration.

By undergoing kinematic analysis, we can compute critical values such as velocity, which are key to understanding subsequent phases of projectile motion.

Inclined Plane

An inclined plane is a sloped surface, allowing an object to be elevated by applying force along the slope. In physics, it is a simple machine, used to lift or lower loads with less force applied along the incline than if lifted directly vertically.
In this problem, the rocket moves on an incline that rises at an angle of 35 degrees. The angle of inclination affects both horizontal and vertical components of motion, critical for analyzing rocket launches.

• **Incline Angle**: Influences the separation of forces and velocities into axes parallel and perpendicular to the surface.
• **Incline Height**: Related directly to the overall height achieved, since the vertical height attained will influence total altitude.
• **Initial Acceleration**: Usually constant, affected by the thrust force applied along the incline.

Understanding inclined planes is pivotal in physics as it showcases how angles affect the decomposition of forces and resultant motions.

Velocity Decomposition

Velocity decomposition is the method of breaking down a vector into its horizontal and vertical components. This process is essential in projectile motion as it allows the analysis of motion across two dimensions separately.
For the rocket problem, once the velocity at the end of the incline is determined, decomposing it into horizontal and vertical components enables better trajectory analysis.

• **Horizontal Component ( v_x )**: Calculated using \( v_x = v \cos(\theta) \), indicates how far the rocket will travel horizontally.
• **Vertical Component ( v_y )**: Calculated using \( v_y = v \sin(\theta) \), dictates how high the rocket will rise.

Through velocity decomposition, one can predict both the height and range of projectiles by clearly understanding how velocity disperses into two distinct dimensions during flight.

Maximum Height Calculation

To calculate the maximum height a projectile reaches, it’s necessary to understand motion dynamics under gravity with initial upward velocity. This concept maps exactly onto our rocket problem after it exits the incline.
Using the formula \( v_y^2 = u_y^2 + 2a_sy \), wherein acceleration due to gravity is the external force acting, we find that the calculation for maximum height requires acknowledgment of when the upward speed diminishes to zero.

• **Initial Vertical Velocity ( u_y )**: At the end of the incline, must be known to calculate the height increase due to momentum.
• **Gravity’s Effect ( a = -9.8 \, \mathrm{m/s}^2 )**: Works against the initial velocity, slowing ascent until momentary stop at peak height.
• **Height achieved**: Represents the additional distance the rocket climbs after leaving the inclined plane.

Maximum height is a frequent objective in projectile problems because it accounts for how high an object will go before descending. This informs ballistic trajectory and energy considerations in dynamic systems.