Problem 65 - University Physics with Modern Physics

Question

A 76.0-kg rock is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake \(\textbf{(Fig. P3.65).}\) The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25 m below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?

Step-by-Step Solution

Answer

(a) 49.5 m/s; (b) 50 m beyond the dam.

Step 1: Identifying Given Values and Required Quantities

We are given the weight of the rock, 76 kg, the height of the cliff, 20 m, the horizontal distance to the dam, 100 m, and the height of the plain, which is 25 m below the dam. We need to find the minimum initial speed (a) such that the rock just barely clears the top of the dam, and (b) the distance from the foot of the dam to where the rock hits the plain.

Step 2: Analyzing the Vertical Motion

Using the kinematic equation for free fall, \( h = \frac{1}{2}gt^2 \), where \( h \) is the height (20 m in this case), and \( g \) is the acceleration due to gravity (9.8 m/s²). Solving for time \( t \) to fall 20 m gives us: \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{9.8}} \approx 2.02 \text{ seconds}.\]

Step 3: Analyzing the Horizontal Motion

The horizontal speed \( v_{x} \) needed to reach the dam's base is based on the horizontal distance. Using the formula for horizontal motion: \( d = v_{x} \times t \), where \( d = 100 \text{ m} \), we find: \[ v_{x} = \frac{d}{t} = \frac{100}{2.02} \approx 49.5 \text{ m/s}. \]

Step 4: Projectile Clears the Obstacle

To ensure the rock clears the dam, we ensure \( v_{x} = 49.5 \text{ m/s} \), where the top of the dam is at the water level (0 elevation compared to top of cliff). Since we calculated the equal time it takes to fall, it clears the dam exactly with no need to account for additional vertical clearance.

Step 5: Determine the Impact Point Beyond the Dam

Since rock clears the dam horizontally at time \( t \), in free fall, the new total height from cliff top to plain is 45 m (since plain is an additional 25 m below the top of the dam). Using the new drop height, \[ t_{total} = \sqrt{\frac{2 \cdot 45}{9.8}} \approx 3.03 \text{ seconds}. \] The total horizontal distance when it hits the plain is \[ v_{x} \times t_{total} = 49.5 \times 3.03 \approx 150.0 \text{ m}.\] The rock lands 50 m beyond the dam (150 m - 100 m).

Key Concepts

Kinematics

Kinematics is a branch of physics that focuses on the motion of objects without considering the forces that cause the motion. In projectile motion, kinematics plays a crucial role. We analyze how the rock moves both horizontally and vertically across the path of its trajectory.

Key kinematic equations help us break down the motion into easily understandable parts. These equations relate quantities like displacement, velocity, acceleration, and time. By using these formulas, we can determine how long it takes an object to move from one point to another and what speed is necessary to complete this motion.

In this exercise, kinematic equations allowed us to find the time the rock takes to fall and the initial speed needed to just clear the dam.

Vertical Motion

Vertical motion concerns how an object moves up or down, influenced by gravity. In this case, the rock's vertical movement starts as it rolls off a cliff.

Using the kinematic equation for free fall, we explored vertical motion to find how long it would take the rock to drop 20 meters. The equation is given by: \( h = \frac{1}{2}gt^2 \), where \( h \) is the height, and \( g \) is the acceleration due to gravity, 9.8 m/s\(^2\). By rearranging this equation, we solved for the time \( t \): \[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{9.8}} \approx 2.02 \text{ seconds}.\]
This time indicates how long it takes the rock to hit the water level from the cliff.

Horizontal Motion

While vertical motion dictates how fast the object falls, horizontal motion shows how far it travels. Horizontal velocity remains constant in the absence of air resistance.

Using the horizontal motion equation \( d = v_{x} \times t \), where \( d \) is the distance and \( v_{x} \) is the horizontal velocity, we find the minimum speed necessary to avoid the dam. By solving: \[ v_{x} = \frac{d}{t} = \frac{100}{2.02} \approx 49.5 \text{ m/s}. \]
This speed ensures that the rock has the ability to travel the 100 meters needed to clear the dam. The horizontal motion remains unaffected by gravity, so this speed sustains until it hits the ground.

Free Fall

Free fall describes any motion of an object where gravity is the only acting force. As soon as the rock leaves the cliff, it is in free fall.

The impact of free fall becomes clear when considering the total fall from the cliff to the plain. The additional 25 meters down from the dam level to the plain increases the total drop to 45 meters. This affects the fall time:\[ t_{total} = \sqrt{\frac{2 \cdot 45}{9.8}} \approx 3.03 \text{ seconds}. \]
The rock's total trajectory then allows us to calculate where it will impact the plain. Since it's in horizontal motion during this free fall, the total distance covered horizontally is \[ v_{x} \times t_{total} = 49.5 \times 3.03 \approx 150.0 \text{ m}. \]
Subtract the 100 meters needed to clear the dam, and you see the rock lands 50 meters beyond it.