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Pregunta

Lambert Stanley

04/03/2024 · escuela secundaria

Question 8 of 10 

You are helping with some repairs at home. You drop a hammer and it hits the floor at a speed of 4 feet per second. If the acceleration due to gravity (g) is 32 feet/second \({ } ^ { 2} \), how far above the ground (h) was the hammer when you dropped it? Use the formula: \[v= \sqrt { 2 g h} \]A. 0.25 feet 

B. 1.0 foot 

C. 0.5 feet 

D. 16.0 feet

Responder
expertRespuesta verificada por expertos

Chandler Chen
Specialized Tutor
5.0 (12votos)

A. 0.25 feet

Solución

UpStudy Free Solution:

 

To determine how far above the ground the hammer was when you dropped it, we can use the given formula:

 

\[v = \sqrt { 2 g h} \]

 

Here:

\(v\) is the final velocity (4 feet per second)

\(g\) is the acceleration due to gravity (32 feet per second\(^ 2\))

\(h\) is the height above the ground

 

First, we square both sides of the equation to eliminate the square root:

 

\[v^ 2 = 2 g h\]

 

Next, we substitute the given values into the equation:

 

\[4^ 2 = 2 \cdot 32 \cdot h\]

 

This simplifies to:

 

\[16 = 64h\]

 

Now, we solve for \(h\):

 

\[h = \frac { 16} { 64} = \frac { 1} { 4} \]

 

Thus, the height \(h\) is 0.25 feet.

 

So, the correct answer is:

 

A. 0.25 feet

 

Supplemental Knowledge

 

In physics, the motion of objects under the influence of gravity is a fundamental concept. When an object is dropped from a height, it accelerates towards the ground due to gravity, which is typically denoted as \(g\). The formula \(v = \sqrt { 2gh} \) relates the final velocity (\(v\)) of an object in free fall to its initial height (\(h\)) above the ground and the acceleration due to gravity (\(g\)).

 

To solve for \(h\), we can rearrange this formula:

\[h = \frac { v^ 2} { 2g} \]

 

Understanding kinematic equations and principles like free fall are vital in solving real-world physics issues, but UpStudy makes the concepts simpler with clear explanations and step-by-step solutions.

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