**Physics Problem: Free Fall on a Different Planet**An object falls freely from rest on a planet where the acceleration due to gravity is \(18 \, \text{m/s}^2\).After \(3.9 \, \text{s}\), what will be its speed? Answer in units of \(\text{m/s}\).**Solution Explanation:**To find the speed of the object after \(3.9 \, \text{s}\), use the formula for velocity in free fall:\[ v = u + at \]where:- \( v \) is the final velocity,- \( u \) is the initial velocity (which is 0 since the object starts from rest),- \( a \) is the acceleration (\(18 \, \text{m/s}^2\) in this case),- \( t \) is the time in seconds (\(3.9 \, \text{s}\)).Plugging in the values:\[ v = 0 + (18 \, \text{m/s}^2)(3.9 \, \text{s}) \]\[ v = 70.2 \, \text{m/s} \]Thus, the speed of the object after \(3.9 \, \text{s}\) will be \(70.2 \, \text{m/s}\).

Name: **Physics Problem: Free Fall on a Different Planet**An object falls f
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Description: **Physics Problem: Free Fall on a Different Planet**An object falls freely from rest on a planet where the acceleration due to gravity is \(18 \, \text{m/s}^2\).After \(3.9 \, \text{s}\), what will be its speed? Answer in units of \(\text{m/s}\).**

The kinematic expression to calculate speed at later time isv=u+gt=0+18 ms23.9 s=70.2 ms

Answered: An object falls freely from rest on a…

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An object falls freely from rest on a planet where the acceleration due to gravity is 18 m/s². After 3.9 s, what will be its speed? Answer in units of m/s.

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**Physics Problem: Free Fall on a Different Planet**

An object falls freely from rest on a planet where the acceleration due to gravity is \(18 \, \text{m/s}^2\).

After \(3.9 \, \text{s}\), what will be its speed?
Answer in units of \(\text{m/s}\).

**Solution Explanation:**

To find the speed of the object after \(3.9 \, \text{s}\), use the formula for velocity in free fall:

\[ v = u + at \]

where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is 0 since the object starts from rest),
- \( a \) is the acceleration (\(18 \, \text{m/s}^2\) in this case),
- \( t \) is the time in seconds (\(3.9 \, \text{s}\)).

Plugging in the values:

\[ v = 0 + (18 \, \text{m/s}^2)(3.9 \, \text{s}) \]
\[ v = 70.2 \, \text{m/s} \]

Thus, the speed of the object after \(3.9 \, \text{s}\) will be \(70.2 \, \text{m/s}\).

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