a 2.00 lb rock is dropped from a bridge to the water 50 below.
a. what is the velocity as it hits the water?
b. how long does it take to hit the water?
c. What is its kinetic energy as it hits the H2o?
Ht. = 50 m.?
a. V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*50 = 980
V = 31.3 m/s.
b. V = Vo + g*t = 31.3 m/s.
0 + 9.8t = 31.3
t = 3.19 s.
c. m = 2.0Lb * 0.454kg/Lb = 0.908 kg.
KE = m*V^2/2 = 0.908*31.3^2/2 = 444.8 J.
To solve these questions, we can use the laws of motion and principles of energy conservation. Let's begin:
a. To find the velocity of the rock as it hits the water, we can use the equation for the velocity of a falling object:
v = sqrt(2gh)
Where:
v = velocity
g = acceleration due to gravity (approximately 32 ft/s^2)
h = height or distance fallen
Given:
g = 32 ft/s^2 (acceleration due to gravity)
h = 50 ft (height or distance fallen)
Substituting the values into the equation, we get:
v = sqrt(2 * 32 ft/s^2 * 50 ft)
v = sqrt(3200 ft^2/s^2)
v ≈ 56.57 ft/s
Therefore, the velocity of the rock as it hits the water is approximately 56.57 ft/s.
b. To determine the time it takes for the rock to hit the water, we can use the equation of motion:
h = (1/2)gt^2
Where:
h = height or distance fallen
g = acceleration due to gravity (approximately 32 ft/s^2)
t = time
Given:
h = 50 ft (height or distance fallen)
g = 32 ft/s^2 (acceleration due to gravity)
Substituting the values into the equation, we have:
50 ft = (1/2) * 32 ft/s^2 * t^2
Rearranging the equation to solve for time, we get:
t = sqrt((50 ft * 2) / 32 ft/s^2)
t = sqrt(100 ft / 32 ft/s^2)
t ≈ 1.58 s
Therefore, it takes approximately 1.58 seconds for the rock to hit the water.
c. To calculate the kinetic energy of the rock as it hits the water, we use the equation:
Kinetic Energy (KE) = (1/2)mv^2
Where:
KE = Kinetic Energy
m = mass
v = velocity
Given:
m = 2.00 lb (mass)
v ≈ 56.57 ft/s (velocity)
First, we need to convert the mass from pounds to slugs by dividing by the acceleration due to gravity:
m_slugs = m / g
m_slugs = 2.00 lb / 32.2 ft/s^2
m_slugs ≈ 0.062 lb⋅s^2/ft
Converting the velocity from ft/s to ft^2/s^2:
v_squared = v^2
v_squared = (56.57 ft/s)^2
v_squared ≈ 3199.78 ft^2/s^2
Now, we can calculate the kinetic energy using the converted values:
KE = (1/2) * m_slugs * v_squared
KE = (1/2) * 0.062 lb⋅s^2/ft * 3199.78 ft^2/s^2
Simplifying the equation, we get:
KE ≈ 99.98 lb⋅ft/s^2
Therefore, the kinetic energy of the rock as it hits the water is approximately 99.98 lb⋅ft/s^2.
To answer these questions, we need to use principles from physics, specifically the laws of motion and the concept of gravitational potential energy.
a. To find the velocity as the rock hits the water, we can use the equation for free fall motion:
v = √(2gh)
Where:
v is the velocity
g is the acceleration due to gravity (approximately 32.2 ft/s^2)
h is the height (50 feet)
Substituting the given values into the equation:
v = √(2 * 32.2 ft/s^2 * 50 ft)
v ≈ √(3220 ft^2/s^2)
v ≈ 56.77 ft/s
Therefore, the velocity of the rock as it hits the water is approximately 56.77 ft/s.
b. To find the time it takes for the rock to hit the water, we can use the equation for the time of free fall:
t = √(2h/g)
Substituting the given values:
t = √(2 * 50 ft / 32.2 ft/s^2)
t ≈ √(100 ft / 32.2 ft/s^2)
t ≈ √(3.1056 s^2)
t ≈ 1.76 s
Therefore, it takes approximately 1.76 seconds for the rock to hit the water.
c. To find the kinetic energy of the rock as it hits the water, we can use the equation for kinetic energy:
KE = (1/2) * m * v^2
Where:
KE is the kinetic energy
m is the mass of the rock (2.00 lb, but we need to convert it to kilograms)
v is the velocity
To convert pounds to kilograms, we can use the conversion factor 1 lb = 0.453592 kg.
So, the mass of the rock is:
m = 2.00 lb * 0.453592 kg/lb
m ≈ 0.907185 kg
Substituting the values into the equation:
KE = (1/2) * 0.907185 kg * (56.77 ft/s)^2
KE ≈ 911.8 joules
Therefore, the kinetic energy of the rock as it hits the water is approximately 911.8 joules.