Problem 21 - University Physics with Modern Physics

Question

A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m/s at an angle of 33.0\(^\circ\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_x-t\), and \(v_y-t\) graphs for the motion.

Step-by-Step Solution

Answer

(a) Maximum height is found using the vertical component of velocity. (b) Calculate the speed before impact using energy or kinematic equations. (c) Horizontal range is the distance traveled horizontally during the flight. (d) Motion graphs show linear, parabolic, and horizontal characteristics.

Step 1: Break Down Initial Velocity

First, resolve the initial velocity into horizontal and vertical components. The initial speed of the rock is 30.0 m/s, thrown at an angle of 33.0° above the horizontal. The horizontal component, \(v_{0x}\), is calculated using the cosine function: \[v_{0x} = v_0 \cdot \cos(33°) = 30.0 \cdot \cos(33°)\] The vertical component, \(v_{0y}\), is calculated using the sine function: \[v_{0y} = v_0 \cdot \sin(33°) = 30.0 \cdot \sin(33°)\]

Step 2: Calculate Maximum Height Above the Roof

To find the maximum height above the roof, use the vertical motion equation considering the upward journey only:\[ v_{y}^2 = v_{0y}^2 - 2g h \]Set \(v_y = 0\) at the peak height to solve for \(h\):\[ 0 = v_{0y}^2 - 2g h \]\[ h = \frac{v_{0y}^2}{2g} \]Plug in the values to calculate \(h\):\(h = \frac{(30.0 \cdot \sin(33°))^2}{2 \cdot 9.81}\).

Step 3: Find Total Height from Ground at Maximum

Add the height of the building to the maximum height above the roof to find the total height from the ground.Total height = building height + maximum height above roof: \[15.0 + h = 15.0 + \frac{(30.0 \cdot \sin(33°))^2}{2 \cdot 9.81}\]

Step 4: Determine Speed Before Impact

Apply energy conservation or use equations of motion to find the speed of the rock just before reaching the ground. Use the vertical movement equations, with the initial and total height taken into account:\(v^2 = v_{0y}^2 + 2g(15.0 + h)\).Calculate the final vertical velocity \(v_y\) and then combine with horizontal component to find the speed just before impact:\[v = \sqrt{v_{0x}^2 + v_y^2}\].

Step 5: Calculate Horizontal Range

Calculate the time of flight by analyzing the vertical motion from the roof back to the ground:Use the formula for time considering the motion to the peak and then back down:\(0 = v_{0y}t - \frac{1}{2}gt^2 + H_0\),solve for total time \(t\) when displacement is zero once it has landed.Then find the horizontal distance covered, \(R\), using:\[R = v_{0x} \cdot t\].

Step 6: Draw Motion Graphs

Graphs to draw:- **x-t graph**: Linear, since horizontal motion has constant velocity \(v_{0x}\).- **y-t graph**: Parabolic, illustrating upward, then downward motion.- **v_x-t graph**: Horizontal line, as \(v_{0x}\) is constant.- **v_y-t graph**: Linear decrease to zero at peak, then more steeply downward due to gravity returning to initial conditions before impact.

Key Concepts

Kinematics

Kinematics is a fascinating branch of physics that focuses on the motion of objects without considering the forces that cause this motion. It's all about describing how things move, which involves parameters like displacement, velocity, and acceleration. When analyzing projectile motion, like the rock thrown from a roof, kinematics helps us break down the motion into two perpendicular components: horizontal and vertical. Both components occur simultaneously, but they act independently due to the absence of horizontal forces in ideal projectile motion—only gravity acts downwards. By applying kinematic equations, we can predict various aspects of the rock's flight, like how high it climbs, how far it travels, and at what speed it impacts the ground. This structured approach makes solving complex motion problems much more manageable.

Velocity Components

In projectile motion, understanding velocity components is crucial. Every projectile has an initial velocity ( v_0 ) that can be split into two basic parts: horizontal ( v_{0x} ) and vertical ( v_{0y} ) components. These components are derived from trigonometric functions based on the angle of projection.

• The horizontal component ( v_{0x} ) is calculated using the cosine of the angle: \[ v_{0x} = v_0 \cdot \cos(\theta) \]
• The vertical component ( v_{0y} ) is determined using the sine of the angle: \[ v_{0y} = v_0 \cdot \sin(\theta) \]

The horizontal velocity remains constant throughout the rock's flight since there's typically no air resistance. However, the vertical velocity begins with the initial value ( v_{0y} ) and changes as the object rises and falls due to gravity ( g ). Decomposing the velocity into these components simplifies the analysis of the motion significantly, making it easier to calculate height, range, and time.

Maximum Height

Finding the maximum height a projectile reaches is an exciting aspect of its motion. This height is where the vertical component of the velocity momentarily becomes zero, before gravity pulls the object back down. We can determine the maximum height above the origin by focusing on the vertical motion only.

• Using the equation: \[ v_y^2 = v_{0y}^2 - 2g h \]

Set v_y = 0 at the peak height to solve for h : \[ h = \frac{v_{0y}^2}{2g} \]Plug in known values to find the height above the launch point, and add it to any initial height, like the height of the building from which the rock was thrown. In this problem, it is crucial to remember to add the building's height to find the total height relative to the ground. Maximum height is valuable since it tells us how high the projectile ventures from its starting point.

Horizontal Range

Discovering the horizontal range of a projectile demonstrates its ability to cover distance. Horizontal range refers to how far along the horizontal plane the projectile travels before hitting the ground. This is reached by considering the time of flight and the constant horizontal velocity ( v_{0x} ).

• First, calculate the time of flight using the entire vertical journey: Consisting of upward, peak, and downward motion.
• The formula is: \[ 0 = v_{0y}t - \frac{1}{2}gt^2 + H_0 \]

Solve this for total time t , the duration the rock is in the air. Then use it in the equation for horizontal distance: \[ R = v_{0x} \cdot t \]This approach allows us to determine how far from the base of the building the rock lands. Horizontal range reflects the logical extent of a projectile under certain initial conditions, unaffected by horizontal forces in ideal scenarios like the exercise stated.