Problem 72 - Fundamentals of Physics

Question

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building \(1.60 \mathrm{~s}\) after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground \(6.00 \mathrm{~s}\) after it is launched. In \(\mathrm{SI}\) units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

Step-by-Step Solution

Answer

(a) 15.70 m/s, (b) 12.53 m, (c) 82.19 m.

Step 1: Identify the given information

The problem provides the following details:- Time to reach maximum height, \(t_1 = 1.60\, \text{s}\)- Total time to hit the ground, \(t_{ ext{total}} = 6.00\, \text{s}\)- Acceleration due to gravity, \(g = 9.81\, \text{m/s}^2\) (downward) We must find the initial velocity, the maximum height reached, and the height of the building.

Step 2: Calculate the initial upward velocity

Use the kinematic equation for velocity: \[ v_f = v_i - gt \]At the maximum height, the final velocity \(v_f = 0\). Rearrange the equation to find the initial velocity \(v_i\):\[ v_i = gt_1 \]Substitute the known values:\[ v_i = (9.81 \, \text{m/s}^2) \times (1.60 \, \text{s}) = 15.696 \, \text{m/s} \]

Step 3: Determine the maximum height above the building

Use the kinematic equation for displacement:\[ h = v_i t - \frac{1}{2}gt^2 \]Substitute to find the maximum height reached from the top of the building:\[ h = (15.696 \, \text{m/s}) \times (1.60 \, \text{s}) - \frac{1}{2}(9.81 \, \text{m/s}^2)(1.60 \, \text{s})^2 \]\[ h = 25.1136 \, \text{m} - 12.5824 \, \text{m} = 12.5312 \, \text{m} \]

Step 4: Calculate the height of the building

The time for the rock to fall from the maximum height back down to the ground is \(t_2 = t_{\text{total}} - t_1 = 6.00 \, \text{s} - 1.60 \, \text{s} = 4.40 \, \text{s}\). Use the equation:\[ h_{\text{building}} = \frac{1}{2}g t_2^2 - h \]Substitute and solve:\[ h_{\text{building}} = \frac{1}{2}(9.81 \, \text{m/s}^2)(4.40 \, \text{s})^2 - 12.5312 \, \text{m} \]\[ h_{\text{building}} = 94.722 \, \text{m} - 12.5312 \, \text{m} = 82.1908 \, \text{m} \]

Step 5: Compile the final answers

(a) The initial upward velocity of the rock is \(15.70 \, \text{m/s}\) (rounded to two decimal places).(b) The maximum height above the building is \(12.53 \, \text{m}\).(c) The building height is \(82.19 \, \text{m}\).

Key Concepts

Initial Velocity

To find the initial velocity of a rock shot upward from a building, we use a kinematic equation that relates velocity, time, and acceleration. At the rock's highest point, its velocity is zero, as all its initial kinetic energy is converted to potential energy. Understanding this is crucial, as the initial velocity directly impacts how high and how far the rock travels.

Since gravity pulls downwards, we need to use the time it takes to reach the maximum height and solve for initial velocity. Using the equation for velocity: \[ v_f = v_i - gt \] where \(v_f\) is the final velocity (0 m/s at the peak), \(v_i\) is the initial velocity, and \(g\) is the acceleration due to gravity. Rearranging gives \[ v_i = gt_1 \], allowing us to calculate the starting speed. This knowledge helps us predict the motion of objects under gravity's influence.

Maximum Height

The maximum height—the highest point the rock reaches above the building—is determined by the initial velocity and gravity's force decelerating the rock upwards. While ascending, the rock loses speed until it momentarily stops at the highest point.

To calculate this, use the kinematic equation for displacement:\[ h = v_i t - \frac{1}{2}gt^2 \].This formula incorporates the initial velocity over time, subtracting the gravitational pull's effect, represented by the term \(\frac{1}{2}gt^2\). In our example, using the values leads to a maximum height of \(12.53 \text{ m}\) above the building, illuminating how energy conservation and gravitational attraction interplay in projectile motion.

Building Height

Identifying the building's height involves a separate calculation, using the total flight time and deductions based on the time spent falling back from the rock's peak. The logic relies on separating the journey into two parts: the ascent to the maximum height and the descent to the ground.

Given that the total time is known, find the time of descent by subtracting the ascent time from it. For the descent, apply \[ h_{building} = \frac{1}{2}gt_2^2 - h \]. This equation deciphers the building’s height by considering the rock's fall trajectory, factoring in gravity's influence and subtracting the height above the building it ascended.

Gravity

Gravity plays a central role in kinematics, acting as the constant force that influences the rock's travel after it is launched. Always directed downward, the gravitational acceleration changes the rock's velocity as it rises and falls.

Represented by the symbol \( g \) and typically approximated as \(9.81 \, \text{m/s}^2\) on Earth's surface, gravity reduces upwards velocity until it reaches zero at the top. Then it accelerates the rock downwards, impacting the overall trajectory and time in the air.

• Gravity is the crucial decelerating force during ascent.
• It is the accelerating force during descent.
• It provides consistent acceleration toward the Earth's surface.

Understanding how gravity affects motion allows for precise predictions of projectile paths and impacts.

Kinematic Equations

Kinematic equations are fundamental tools for analyzing motion. They combine variables like velocity, time, displacement, and acceleration to solve problems in motion, as seen in our rock scenario.

These equations help connect various motion parameters, allowing us to predict one dimension's value when others are known. For instance, in our exercise:

• Determine initial velocity with\( v_f = v_i - gt \).
• Find maximum height with displacement\( h = v_i t - \frac{1}{2}gt^2 \).
• Calculate building height by combining time and gravitational effect.

Each equation illuminates a different facet of motion, essential for comprehensively understanding the dynamics of an object's journey.