# A stone is dropped from the top of a cliff. It hits the ground below after 3.25 s. How high is the cliff?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

15
Nov 11, 2015

Use the formula $d = \frac{1}{2} a {t}^{2}$. This formula is valid for constant acceleration and no initial velocity. Use acceleration due to gravity$= g = 9.8 \frac{m}{s} ^ 2$, and $t = 3.25 s$. Then $d = \frac{1}{2} \left(9.8 \frac{m}{s} ^ 2\right) {\left(3.25 s\right)}^{2} = 51.8 m$.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Nov 11, 2015

$51.8 m$

#### Explanation:

Lets list our variables first
Deltay = ?m This is what we are trying to find
${v}_{i} = 0 \frac{m}{s}$ The initial velocity is 0 because our object dropped from rest
${v}_{f} = 0$ The final velocity is not relevant in this problem
$t = 3.25 \sec$ We are given the time in seconds
$a = 9.80 \frac{m}{s} ^ 2$ This is the downward acceleration due to gravity

Now we use our kinematic equation
$\Delta y = {v}_{i} \cdot t + \frac{a \cdot {t}^{2}}{2}$
Plug in the variables

$\Delta y = 0 \frac{m}{s} \cdot 3.25 \sec + \frac{9.80 \frac{m}{\cancel{{s}^{2}}} \cdot {3.25}^{2} \cancel{{\sec}^{2}}}{2}$

$\Delta y = 51.75 m$

Because there are 3 significant figures, we round the answer to $51.8 m$

• 7 minutes ago
• 10 minutes ago
• 12 minutes ago
• 15 minutes ago
• 5 minutes ago
• 5 minutes ago
• 5 minutes ago
• 6 minutes ago
• 6 minutes ago
• 7 minutes ago
• 7 minutes ago
• 10 minutes ago
• 12 minutes ago
• 15 minutes ago